Four married couples attend a party. Each person shakes hands with every other person, except their own...
$begingroup$
Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
My book gave the answer as $24$. I do not understand why.
I thought of it like this:
You have four pairs of couples, so you can think of it as
M1W2, M2W2, M3W3, M4W4,
where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4times 6$ handshakes, but in my answer, you are double counting.
How do I approach this problem?
combinatorics
edited 7 secs ago
Jack
27.5k1782202
asked yesterday
ZakuZaku
1245
$endgroup$
put on hold as off-topic by Xander Henderson, RRL, Lee David Chung Lin, Song, abc... 20 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Song, abc...
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 3 more comments
$begingroup$
Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
My book gave the answer as $24$. I do not understand why.
I thought of it like this:
You have four pairs of couples, so you can think of it as
M1W2, M2W2, M3W3, M4W4,
where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4times 6$ handshakes, but in my answer, you are double counting.
How do I approach this problem?
combinatorics
edited 7 secs ago
Jack
27.5k1782202
asked yesterday
ZakuZaku
1245
$endgroup$
put on hold as off-topic by Xander Henderson, RRL, Lee David Chung Lin, Song, abc... 20 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Song, abc...
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
yesterday
1
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
yesterday
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
yesterday
4
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
yesterday
2
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
yesterday
|
show 3 more comments
$begingroup$
Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
My book gave the answer as $24$. I do not understand why.
I thought of it like this:
You have four pairs of couples, so you can think of it as
M1W2, M2W2, M3W3, M4W4,
where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4times 6$ handshakes, but in my answer, you are double counting.
How do I approach this problem?
combinatorics
edited 7 secs ago
Jack
27.5k1782202
asked yesterday
ZakuZaku
1245
$endgroup$
Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
My book gave the answer as $24$. I do not understand why.
I thought of it like this:
You have four pairs of couples, so you can think of it as
M1W2, M2W2, M3W3, M4W4,
where
M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4times 6$ handshakes, but in my answer, you are double counting.
How do I approach this problem?
combinatorics
combinatorics
edited 7 secs ago
Jack
27.5k1782202
asked yesterday
ZakuZaku
1245
edited 7 secs ago
Jack
27.5k1782202
asked yesterday
ZakuZaku
1245
edited 7 secs ago
Jack
27.5k1782202
edited 7 secs ago
Jack
27.5k1782202
edited 7 secs ago
Jack
27.5k1782202
27.5k1782202
asked yesterday
ZakuZaku
1245
asked yesterday
ZakuZaku
1245
asked yesterday
ZakuZaku
1245
1245
put on hold as off-topic by Xander Henderson, RRL, Lee David Chung Lin, Song, abc... 20 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Song, abc...
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Xander Henderson, RRL, Lee David Chung Lin, Song, abc... 20 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Song, abc...
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
yesterday
1
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
yesterday
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
yesterday
4
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
yesterday
2
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
yesterday
|
show 3 more comments
2
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
yesterday
1
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
yesterday
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
yesterday
4
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
yesterday
2
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
yesterday
2
2
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
yesterday
$begingroup$
In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
$endgroup$
– M. Vinay
yesterday
1
1
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
yesterday
$begingroup$
And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
$endgroup$
– M. Vinay
yesterday
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
yesterday
$begingroup$
I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
$endgroup$
– DanielV
yesterday
4
4
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
yesterday
$begingroup$
@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
$endgroup$
– M. Vinay
yesterday
2
2
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
yesterday
$begingroup$
Possible duplicate of Handshakes in a party
$endgroup$
– Xander Henderson
yesterday
|
show 3 more comments
8 Answers
8
active
oldest
votes
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$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered yesterday
fleabloodfleablood
72.8k22788
$endgroup$
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered yesterday
Austin MohrAustin Mohr
20.7k35199
$endgroup$
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
yesterday
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
19 hours ago
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered yesterday
trancelocationtrancelocation
12.7k1827
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add a comment |
$begingroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered yesterday
dodidodi
692
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dodi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
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It shouldn't be 3! [(n-1)!]. It should be n(n-1)/2. Try using a different value for n and see what happens.
$endgroup$
– CJ Dennis
21 hours ago
add a comment |
$begingroup$
Each line is a handshake between the required two people. There are 24 lines:
answered yesterday
Witness Protection ID 44583292Witness Protection ID 44583292
23113
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add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered yesterday
beefstew2011beefstew2011
687
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1
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Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
yesterday
add a comment |
$begingroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered yesterday
Vijendra ParasharVijendra Parashar
191
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5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
yesterday
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
yesterday
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
13 hours ago
add a comment |
$begingroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered yesterday
Chand16Chand16
255
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add a comment |
8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered yesterday
fleabloodfleablood
72.8k22788
$endgroup$
add a comment |
$begingroup$
$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered yesterday
fleabloodfleablood
72.8k22788
$endgroup$
add a comment |
$begingroup$
$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered yesterday
fleabloodfleablood
72.8k22788
$endgroup$
$8$ people. Each experiences handshakes with $6$ people. There are $6times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48div 2=24$ handshakes.
answered yesterday
fleabloodfleablood
72.8k22788
answered yesterday
fleabloodfleablood
72.8k22788
answered yesterday
fleabloodfleablood
72.8k22788
answered yesterday
fleabloodfleablood
72.8k22788
72.8k22788
add a comment |
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered yesterday
Austin MohrAustin Mohr
20.7k35199
$endgroup$
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
yesterday
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
19 hours ago
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered yesterday
Austin MohrAustin Mohr
20.7k35199
$endgroup$
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
yesterday
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
19 hours ago
add a comment |
$begingroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered yesterday
Austin MohrAustin Mohr
20.7k35199
$endgroup$
Suppose the spouses were allowed to shake each other's hands. That would give you $binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
answered yesterday
Austin MohrAustin Mohr
20.7k35199
answered yesterday
Austin MohrAustin Mohr
20.7k35199
answered yesterday
Austin MohrAustin Mohr
20.7k35199
answered yesterday
Austin MohrAustin Mohr
20.7k35199
20.7k35199
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
yesterday
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
19 hours ago
add a comment |
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
yesterday
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
19 hours ago
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
yesterday
$begingroup$
This uses Inclusion-Exclusion Principle.
$endgroup$
– smci
yesterday
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
19 hours ago
$begingroup$
Inclusion-Exclusion helps to find the cardinality of a union of non-disjoint sets. I'm merely using the fact that a set together with its complement (which are disjoint) comprise the entire universe.
$endgroup$
– Austin Mohr
19 hours ago
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered yesterday
trancelocationtrancelocation
12.7k1827
$endgroup$
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered yesterday
trancelocationtrancelocation
12.7k1827
$endgroup$
add a comment |
$begingroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered yesterday
trancelocationtrancelocation
12.7k1827
$endgroup$
You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $color{blue}{binom{8}{2}}$
- Number of pairs who do not shake hands: $color{blue}{4}$
It follows:
$$mbox{number of hand shakes without pairs} = color{blue}{binom{8}{2}} - color{blue}{4} = frac{8cdot 7}{2} - 4 = 24$$
answered yesterday
trancelocationtrancelocation
12.7k1827
answered yesterday
trancelocationtrancelocation
12.7k1827
answered yesterday
trancelocationtrancelocation
12.7k1827
answered yesterday
trancelocationtrancelocation
12.7k1827
12.7k1827
add a comment |
add a comment |
$begingroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered yesterday
dodidodi
692
New contributor
dodi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
It shouldn't be 3! [(n-1)!]. It should be n(n-1)/2. Try using a different value for n and see what happens.
$endgroup$
– CJ Dennis
21 hours ago
add a comment |
$begingroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered yesterday
dodidodi
692
New contributor
dodi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
It shouldn't be 3! [(n-1)!]. It should be n(n-1)/2. Try using a different value for n and see what happens.
$endgroup$
– CJ Dennis
21 hours ago
add a comment |
$begingroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered yesterday
dodidodi
692
New contributor
dodi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is
$sum_{k=1}^{n-1}k=frac{n(n-1)}{2}$.
answered yesterday
dodidodi
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edited 12 hours ago
answered yesterday
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dodidodi
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answered yesterday
dodidodi
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2
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It shouldn't be 3! [(n-1)!]. It should be n(n-1)/2. Try using a different value for n and see what happens.
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– CJ Dennis
21 hours ago
add a comment |
2
$begingroup$
It shouldn't be 3! [(n-1)!]. It should be n(n-1)/2. Try using a different value for n and see what happens.
$endgroup$
– CJ Dennis
21 hours ago
2
2
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It shouldn't be 3! [(n-1)!]. It should be n(n-1)/2. Try using a different value for n and see what happens.
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– CJ Dennis
21 hours ago
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It shouldn't be 3! [(n-1)!]. It should be n(n-1)/2. Try using a different value for n and see what happens.
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– CJ Dennis
21 hours ago
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Each line is a handshake between the required two people. There are 24 lines:
answered yesterday
Witness Protection ID 44583292Witness Protection ID 44583292
23113
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add a comment |
$begingroup$
Each line is a handshake between the required two people. There are 24 lines:
answered yesterday
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
add a comment |
$begingroup$
Each line is a handshake between the required two people. There are 24 lines:
answered yesterday
Witness Protection ID 44583292Witness Protection ID 44583292
23113
$endgroup$
Each line is a handshake between the required two people. There are 24 lines:
answered yesterday
Witness Protection ID 44583292Witness Protection ID 44583292
23113
answered yesterday
Witness Protection ID 44583292Witness Protection ID 44583292
23113
answered yesterday
Witness Protection ID 44583292Witness Protection ID 44583292
23113
answered yesterday
Witness Protection ID 44583292Witness Protection ID 44583292
23113
23113
add a comment |
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered yesterday
beefstew2011beefstew2011
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1
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Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
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– M. Vinay
yesterday
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered yesterday
beefstew2011beefstew2011
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1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
yesterday
add a comment |
$begingroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered yesterday
beefstew2011beefstew2011
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$endgroup$
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$sum_{i=1}^k (4k-4i) = sum_{i=1}^k4k - sum_{i=1}^k4i = 4k^2 - 4frac{k(k+1)}{2} = 4(k^2 - frac{k^2+k}{2}) = 4(k^2 - (frac{k^2}{2} + frac{k}{2})) = 4(frac{k^2}{2}-frac{k}{2}) = 2(k^2-k)$$
for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.
answered yesterday
beefstew2011beefstew2011
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edited yesterday
answered yesterday
beefstew2011beefstew2011
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answered yesterday
beefstew2011beefstew2011
687
answered yesterday
beefstew2011beefstew2011
687
687
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1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
yesterday
add a comment |
1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
yesterday
1
1
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
yesterday
$begingroup$
Well… Each of the $2k$ people shakes hands with $2k - 1 - 1 = 2k - 2$ others (everyone except the spouse). So that's $2k(2k- 2) = 4k(k - 1)$, but since every handshake must've been counted twice, divide that by $2$ to get $2k(k - 1)$ handshakes in total.
$endgroup$
– M. Vinay
yesterday
add a comment |
$begingroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered yesterday
Vijendra ParasharVijendra Parashar
191
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5
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How is different from fleablood's answer?
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– Toby Mak
yesterday
1
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@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
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– Vijendra Parashar
yesterday
1
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I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
13 hours ago
add a comment |
$begingroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered yesterday
Vijendra ParasharVijendra Parashar
191
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$endgroup$
5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
yesterday
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
yesterday
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
13 hours ago
add a comment |
$begingroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered yesterday
Vijendra ParasharVijendra Parashar
191
New contributor
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$endgroup$
A simple approach:
There are 8 person in total.
Each one will shake hands with 6 others.
Total shakehands from individual perspective: 6*8 gives 48
Actual shakehands: 48/2 = 24
answered yesterday
Vijendra ParasharVijendra Parashar
191
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answered yesterday
Vijendra ParasharVijendra Parashar
191
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answered yesterday
Vijendra ParasharVijendra Parashar
191
answered yesterday
Vijendra ParasharVijendra Parashar
191
191
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5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
yesterday
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
yesterday
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
13 hours ago
add a comment |
5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
yesterday
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
yesterday
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
13 hours ago
5
5
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
yesterday
$begingroup$
How is different from fleablood's answer?
$endgroup$
– Toby Mak
yesterday
1
1
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
yesterday
$begingroup$
@TobyMak sorry, I really didn't see it. When I posted there were only four answers including mine. That answer was really not there, completely surprised. I don't know how this happened?
$endgroup$
– Vijendra Parashar
yesterday
1
1
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
13 hours ago
$begingroup$
I see. Since you wrote your answer independently from fleablood, it's only fair to keep your answer.
$endgroup$
– Toby Mak
13 hours ago
add a comment |
$begingroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered yesterday
Chand16Chand16
255
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add a comment |
$begingroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered yesterday
Chand16Chand16
255
$endgroup$
add a comment |
$begingroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered yesterday
Chand16Chand16
255
$endgroup$
If all of them handshakes each other then there are 8!/2! =28 handshakes, but none of them handshake with their own spouse so their are 28-4=24 handshakes.
answered yesterday
Chand16Chand16
255
answered yesterday
Chand16Chand16
255
answered yesterday
Chand16Chand16
255
answered yesterday
Chand16Chand16
255
255
add a comment |
add a comment |
2
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In your answer, you both overcounted and undercounted, and incidentally these happened to cancel out and give you the correct answer without having to do anything further. You did $4 times (text{Handshakes done by the men})$, which overcounted the man-man handshakes, but left out the woman-woman handshakes.
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– M. Vinay
yesterday
1
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And that's easily fixed by counting all such handshakes in the same way, not just those done by men, so you get $48$. And now, as you said, you have indeed double-counted. But if you know it's exactly double counting, you can get the answer by halving it!
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– M. Vinay
yesterday
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I recommend when you have a problem like this you can't solve, try solving an easier version first, like only 2 couples and anything goes.
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– DanielV
yesterday
4
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@Issel No, Person #2 being the spouse of Person #1, also has to shake hands with $6$ people, and so on, so it's $6 + 6 + 4 + 4 + 2 + 2 + 0 + 0 = 24$.
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– M. Vinay
yesterday
2
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Possible duplicate of Handshakes in a party
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– Xander Henderson
yesterday