Counting Permutation of a set with the condition that $a_1 < a_2$
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I'm not sure how the answer is a. How I approached this is:
For $a_1$ < $a_2$ we can select $a_2$ to be $n$ (1 way) and then we have no choices for $a_1$ because it has to be less than $a_2$. Or actually, then it should be $n-1$. Then the remaining ways to arrange it can be done in $(n-2)!$ ways.
So, I get: $n×(n-1)×(n-2)!$ as the total permutations. Now, what I realized it is that what I got and options b and care actually the same thing surprisingly! So with that intuition, option a has to be right. But I really don't get why.
Like if I take a set with {$1,2,3,4$} then I pick $a_2$ to be $4$, my options for $a_1$ will be $1$,$2$ or $3$. But I have it as $n-1$ which will only give me $3$. I also tried $frac{n-1}{2}$ but that would not give me an integer for some cases. I'm just so confused on how to get an expression that makes sense to me. I
combinatorics permutations
edited Dec 13 '18 at 4:27
Neil hawking
1
asked Dec 13 '18 at 2:29
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
I'm not sure how the answer is a. How I approached this is:
For $a_1$ < $a_2$ we can select $a_2$ to be $n$ (1 way) and then we have no choices for $a_1$ because it has to be less than $a_2$. Or actually, then it should be $n-1$. Then the remaining ways to arrange it can be done in $(n-2)!$ ways.
So, I get: $n×(n-1)×(n-2)!$ as the total permutations. Now, what I realized it is that what I got and options b and care actually the same thing surprisingly! So with that intuition, option a has to be right. But I really don't get why.
Like if I take a set with {$1,2,3,4$} then I pick $a_2$ to be $4$, my options for $a_1$ will be $1$,$2$ or $3$. But I have it as $n-1$ which will only give me $3$. I also tried $frac{n-1}{2}$ but that would not give me an integer for some cases. I'm just so confused on how to get an expression that makes sense to me. I
combinatorics permutations
edited Dec 13 '18 at 4:27
Neil hawking
1
asked Dec 13 '18 at 2:29
TobyToby
1577
$endgroup$
add a comment |
$begingroup$
I'm not sure how the answer is a. How I approached this is:
For $a_1$ < $a_2$ we can select $a_2$ to be $n$ (1 way) and then we have no choices for $a_1$ because it has to be less than $a_2$. Or actually, then it should be $n-1$. Then the remaining ways to arrange it can be done in $(n-2)!$ ways.
So, I get: $n×(n-1)×(n-2)!$ as the total permutations. Now, what I realized it is that what I got and options b and care actually the same thing surprisingly! So with that intuition, option a has to be right. But I really don't get why.
Like if I take a set with {$1,2,3,4$} then I pick $a_2$ to be $4$, my options for $a_1$ will be $1$,$2$ or $3$. But I have it as $n-1$ which will only give me $3$. I also tried $frac{n-1}{2}$ but that would not give me an integer for some cases. I'm just so confused on how to get an expression that makes sense to me. I
combinatorics permutations
edited Dec 13 '18 at 4:27
Neil hawking
1
asked Dec 13 '18 at 2:29
TobyToby
1577
$endgroup$
I'm not sure how the answer is a. How I approached this is:
For $a_1$ < $a_2$ we can select $a_2$ to be $n$ (1 way) and then we have no choices for $a_1$ because it has to be less than $a_2$. Or actually, then it should be $n-1$. Then the remaining ways to arrange it can be done in $(n-2)!$ ways.
So, I get: $n×(n-1)×(n-2)!$ as the total permutations. Now, what I realized it is that what I got and options b and care actually the same thing surprisingly! So with that intuition, option a has to be right. But I really don't get why.
Like if I take a set with {$1,2,3,4$} then I pick $a_2$ to be $4$, my options for $a_1$ will be $1$,$2$ or $3$. But I have it as $n-1$ which will only give me $3$. I also tried $frac{n-1}{2}$ but that would not give me an integer for some cases. I'm just so confused on how to get an expression that makes sense to me. I
combinatorics permutations
combinatorics permutations
edited Dec 13 '18 at 4:27
Neil hawking
1
asked Dec 13 '18 at 2:29
TobyToby
1577
edited Dec 13 '18 at 4:27
Neil hawking
1
asked Dec 13 '18 at 2:29
TobyToby
1577
edited Dec 13 '18 at 4:27
Neil hawking
1
edited Dec 13 '18 at 4:27
Neil hawking
1
edited Dec 13 '18 at 4:27
Neil hawking
1
1
asked Dec 13 '18 at 2:29
TobyToby
1577
asked Dec 13 '18 at 2:29
TobyToby
1577
asked Dec 13 '18 at 2:29
TobyToby
1577
1577
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Think about it this way:
There are $n!$ permutations without any restrictions, and by symmetry there are just as many permutations where the first element is smaller than the second, as there are permutations where the first element is greater than the second.
Or: for every permutations with $a_1 < a_2$ you can get the same permutation but with $a_1$ and $a_2$ swapped, and thus one where $a_2 < a_1$, and vice versa. So, for every one where $a_1<a_2$ you have one where $a_2<a_1$, so the ones where $a_1<a_2$ is half of all of them.
answered Dec 13 '18 at 2:34
Bram28Bram28
63.2k44793
$endgroup$
$begingroup$
I still don't understand how there can be just as many permutations for both. If I have a set say {1,2,3,4,5}, how will what you're saying work?
$endgroup$
– Toby
Dec 13 '18 at 2:41
$begingroup$
@Toby OK, you have $6$ permutations that start with $1,2$ ... but also $6$ that start with $2,1$. Then, you have $6$ permutations that start with $1,3$ ... and $6$ that start with $3,1$. And for the $6$ that start with $2,4$, you have $6$ that start with $4,2$, ... etc. Do you see it now?
$endgroup$
– Bram28
Dec 13 '18 at 2:46
$begingroup$
OK, so this is saying that the first 2 elements will always have the same permutation?
$endgroup$
– Toby
Dec 13 '18 at 2:51
$begingroup$
@Toby Not quite .. it is saying that, given whatever the first two elements are, those two can either be put in the order where the first is smaller than the second, or the other way around, however the rest of the humbers are permuated after that. So, just as you have $1,2,4,3,5$, you also have $2,1,4,3,5$. Or, just as you have $3,5,1,2,4$, you have $5,3,1,2,4$. So they all come in pairs like at, where the sequence after the first two is exactly the same.
$endgroup$
– Bram28
Dec 13 '18 at 2:57
1
$begingroup$
@Toby Right, there are $(n-2)!$ ways to permuate the $n-2$ numbers after the first two. And yes, there are $n choose 2$ ways to pick the first two. But what I am saying is: it really doesn't matter what the exact number of permutations is or what the number of ways to pick the first two numbers is ... all you need to do is realize that whatever those numbers are with $a_1<a_2$, you get the same numbers for $a_2<a_1$ ... so each is half of all $n!$ permutations. No complicated calculations needed!
$endgroup$
– Bram28
Dec 13 '18 at 3:26
|
show 1 more comment
$begingroup$
The condition is only on the first 2, and not on the rest. Once the first two satisfy your requirement the reamining can be arranged in any way, that is $(n-2)!$ Now the first two have are two elements from $n$ with specific order (ascending), and no rearrangement which is ${nchoose 2}$. So it is the product of these two numbers, ${nchoose 2}times (n-2)!$ which is same as $frac12 n!$
answered Dec 13 '18 at 2:42
P VanchinathanP Vanchinathan
15.2k12136
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Why is it nchoose2 for the first 2 elements?
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– Toby
Dec 13 '18 at 3:07
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This is the first lesson in counting. The number of ways of picking r elements (and not rearranging them) is n-choose-r. Not rearranging them can be guaranteed if we agree to stick to ascending order arrangement as the only way.
$endgroup$
– P Vanchinathan
Dec 13 '18 at 3:14
add a comment |
$begingroup$
The number of all permutation is $n!$. Half of it with $a_1<a_2$ and the other half with $a_2<a_1$.
answered Dec 13 '18 at 2:36
user9077user9077
1,239612
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Think about it this way:
There are $n!$ permutations without any restrictions, and by symmetry there are just as many permutations where the first element is smaller than the second, as there are permutations where the first element is greater than the second.
Or: for every permutations with $a_1 < a_2$ you can get the same permutation but with $a_1$ and $a_2$ swapped, and thus one where $a_2 < a_1$, and vice versa. So, for every one where $a_1<a_2$ you have one where $a_2<a_1$, so the ones where $a_1<a_2$ is half of all of them.
answered Dec 13 '18 at 2:34
Bram28Bram28
63.2k44793
$endgroup$
$begingroup$
I still don't understand how there can be just as many permutations for both. If I have a set say {1,2,3,4,5}, how will what you're saying work?
$endgroup$
– Toby
Dec 13 '18 at 2:41
$begingroup$
@Toby OK, you have $6$ permutations that start with $1,2$ ... but also $6$ that start with $2,1$. Then, you have $6$ permutations that start with $1,3$ ... and $6$ that start with $3,1$. And for the $6$ that start with $2,4$, you have $6$ that start with $4,2$, ... etc. Do you see it now?
$endgroup$
– Bram28
Dec 13 '18 at 2:46
$begingroup$
OK, so this is saying that the first 2 elements will always have the same permutation?
$endgroup$
– Toby
Dec 13 '18 at 2:51
$begingroup$
@Toby Not quite .. it is saying that, given whatever the first two elements are, those two can either be put in the order where the first is smaller than the second, or the other way around, however the rest of the humbers are permuated after that. So, just as you have $1,2,4,3,5$, you also have $2,1,4,3,5$. Or, just as you have $3,5,1,2,4$, you have $5,3,1,2,4$. So they all come in pairs like at, where the sequence after the first two is exactly the same.
$endgroup$
– Bram28
Dec 13 '18 at 2:57
1
$begingroup$
@Toby Right, there are $(n-2)!$ ways to permuate the $n-2$ numbers after the first two. And yes, there are $n choose 2$ ways to pick the first two. But what I am saying is: it really doesn't matter what the exact number of permutations is or what the number of ways to pick the first two numbers is ... all you need to do is realize that whatever those numbers are with $a_1<a_2$, you get the same numbers for $a_2<a_1$ ... so each is half of all $n!$ permutations. No complicated calculations needed!
$endgroup$
– Bram28
Dec 13 '18 at 3:26
|
show 1 more comment
$begingroup$
Think about it this way:
There are $n!$ permutations without any restrictions, and by symmetry there are just as many permutations where the first element is smaller than the second, as there are permutations where the first element is greater than the second.
Or: for every permutations with $a_1 < a_2$ you can get the same permutation but with $a_1$ and $a_2$ swapped, and thus one where $a_2 < a_1$, and vice versa. So, for every one where $a_1<a_2$ you have one where $a_2<a_1$, so the ones where $a_1<a_2$ is half of all of them.
answered Dec 13 '18 at 2:34
Bram28Bram28
63.2k44793
$endgroup$
$begingroup$
I still don't understand how there can be just as many permutations for both. If I have a set say {1,2,3,4,5}, how will what you're saying work?
$endgroup$
– Toby
Dec 13 '18 at 2:41
$begingroup$
@Toby OK, you have $6$ permutations that start with $1,2$ ... but also $6$ that start with $2,1$. Then, you have $6$ permutations that start with $1,3$ ... and $6$ that start with $3,1$. And for the $6$ that start with $2,4$, you have $6$ that start with $4,2$, ... etc. Do you see it now?
$endgroup$
– Bram28
Dec 13 '18 at 2:46
$begingroup$
OK, so this is saying that the first 2 elements will always have the same permutation?
$endgroup$
– Toby
Dec 13 '18 at 2:51
$begingroup$
@Toby Not quite .. it is saying that, given whatever the first two elements are, those two can either be put in the order where the first is smaller than the second, or the other way around, however the rest of the humbers are permuated after that. So, just as you have $1,2,4,3,5$, you also have $2,1,4,3,5$. Or, just as you have $3,5,1,2,4$, you have $5,3,1,2,4$. So they all come in pairs like at, where the sequence after the first two is exactly the same.
$endgroup$
– Bram28
Dec 13 '18 at 2:57
1
$begingroup$
@Toby Right, there are $(n-2)!$ ways to permuate the $n-2$ numbers after the first two. And yes, there are $n choose 2$ ways to pick the first two. But what I am saying is: it really doesn't matter what the exact number of permutations is or what the number of ways to pick the first two numbers is ... all you need to do is realize that whatever those numbers are with $a_1<a_2$, you get the same numbers for $a_2<a_1$ ... so each is half of all $n!$ permutations. No complicated calculations needed!
$endgroup$
– Bram28
Dec 13 '18 at 3:26
|
show 1 more comment
$begingroup$
Think about it this way:
There are $n!$ permutations without any restrictions, and by symmetry there are just as many permutations where the first element is smaller than the second, as there are permutations where the first element is greater than the second.
Or: for every permutations with $a_1 < a_2$ you can get the same permutation but with $a_1$ and $a_2$ swapped, and thus one where $a_2 < a_1$, and vice versa. So, for every one where $a_1<a_2$ you have one where $a_2<a_1$, so the ones where $a_1<a_2$ is half of all of them.
answered Dec 13 '18 at 2:34
Bram28Bram28
63.2k44793
$endgroup$
Think about it this way:
There are $n!$ permutations without any restrictions, and by symmetry there are just as many permutations where the first element is smaller than the second, as there are permutations where the first element is greater than the second.
Or: for every permutations with $a_1 < a_2$ you can get the same permutation but with $a_1$ and $a_2$ swapped, and thus one where $a_2 < a_1$, and vice versa. So, for every one where $a_1<a_2$ you have one where $a_2<a_1$, so the ones where $a_1<a_2$ is half of all of them.
answered Dec 13 '18 at 2:34
Bram28Bram28
63.2k44793
edited Dec 13 '18 at 2:43
answered Dec 13 '18 at 2:34
Bram28Bram28
63.2k44793
answered Dec 13 '18 at 2:34
Bram28Bram28
63.2k44793
answered Dec 13 '18 at 2:34
Bram28Bram28
63.2k44793
63.2k44793
$begingroup$
I still don't understand how there can be just as many permutations for both. If I have a set say {1,2,3,4,5}, how will what you're saying work?
$endgroup$
– Toby
Dec 13 '18 at 2:41
$begingroup$
@Toby OK, you have $6$ permutations that start with $1,2$ ... but also $6$ that start with $2,1$. Then, you have $6$ permutations that start with $1,3$ ... and $6$ that start with $3,1$. And for the $6$ that start with $2,4$, you have $6$ that start with $4,2$, ... etc. Do you see it now?
$endgroup$
– Bram28
Dec 13 '18 at 2:46
$begingroup$
OK, so this is saying that the first 2 elements will always have the same permutation?
$endgroup$
– Toby
Dec 13 '18 at 2:51
$begingroup$
@Toby Not quite .. it is saying that, given whatever the first two elements are, those two can either be put in the order where the first is smaller than the second, or the other way around, however the rest of the humbers are permuated after that. So, just as you have $1,2,4,3,5$, you also have $2,1,4,3,5$. Or, just as you have $3,5,1,2,4$, you have $5,3,1,2,4$. So they all come in pairs like at, where the sequence after the first two is exactly the same.
$endgroup$
– Bram28
Dec 13 '18 at 2:57
1
$begingroup$
@Toby Right, there are $(n-2)!$ ways to permuate the $n-2$ numbers after the first two. And yes, there are $n choose 2$ ways to pick the first two. But what I am saying is: it really doesn't matter what the exact number of permutations is or what the number of ways to pick the first two numbers is ... all you need to do is realize that whatever those numbers are with $a_1<a_2$, you get the same numbers for $a_2<a_1$ ... so each is half of all $n!$ permutations. No complicated calculations needed!
$endgroup$
– Bram28
Dec 13 '18 at 3:26
|
show 1 more comment
$begingroup$
I still don't understand how there can be just as many permutations for both. If I have a set say {1,2,3,4,5}, how will what you're saying work?
$endgroup$
– Toby
Dec 13 '18 at 2:41
$begingroup$
@Toby OK, you have $6$ permutations that start with $1,2$ ... but also $6$ that start with $2,1$. Then, you have $6$ permutations that start with $1,3$ ... and $6$ that start with $3,1$. And for the $6$ that start with $2,4$, you have $6$ that start with $4,2$, ... etc. Do you see it now?
$endgroup$
– Bram28
Dec 13 '18 at 2:46
$begingroup$
OK, so this is saying that the first 2 elements will always have the same permutation?
$endgroup$
– Toby
Dec 13 '18 at 2:51
$begingroup$
@Toby Not quite .. it is saying that, given whatever the first two elements are, those two can either be put in the order where the first is smaller than the second, or the other way around, however the rest of the humbers are permuated after that. So, just as you have $1,2,4,3,5$, you also have $2,1,4,3,5$. Or, just as you have $3,5,1,2,4$, you have $5,3,1,2,4$. So they all come in pairs like at, where the sequence after the first two is exactly the same.
$endgroup$
– Bram28
Dec 13 '18 at 2:57
1
$begingroup$
@Toby Right, there are $(n-2)!$ ways to permuate the $n-2$ numbers after the first two. And yes, there are $n choose 2$ ways to pick the first two. But what I am saying is: it really doesn't matter what the exact number of permutations is or what the number of ways to pick the first two numbers is ... all you need to do is realize that whatever those numbers are with $a_1<a_2$, you get the same numbers for $a_2<a_1$ ... so each is half of all $n!$ permutations. No complicated calculations needed!
$endgroup$
– Bram28
Dec 13 '18 at 3:26
$begingroup$
I still don't understand how there can be just as many permutations for both. If I have a set say {1,2,3,4,5}, how will what you're saying work?
$endgroup$
– Toby
Dec 13 '18 at 2:41
$begingroup$
I still don't understand how there can be just as many permutations for both. If I have a set say {1,2,3,4,5}, how will what you're saying work?
$endgroup$
– Toby
Dec 13 '18 at 2:41
$begingroup$
@Toby OK, you have $6$ permutations that start with $1,2$ ... but also $6$ that start with $2,1$. Then, you have $6$ permutations that start with $1,3$ ... and $6$ that start with $3,1$. And for the $6$ that start with $2,4$, you have $6$ that start with $4,2$, ... etc. Do you see it now?
$endgroup$
– Bram28
Dec 13 '18 at 2:46
$begingroup$
@Toby OK, you have $6$ permutations that start with $1,2$ ... but also $6$ that start with $2,1$. Then, you have $6$ permutations that start with $1,3$ ... and $6$ that start with $3,1$. And for the $6$ that start with $2,4$, you have $6$ that start with $4,2$, ... etc. Do you see it now?
$endgroup$
– Bram28
Dec 13 '18 at 2:46
$begingroup$
OK, so this is saying that the first 2 elements will always have the same permutation?
$endgroup$
– Toby
Dec 13 '18 at 2:51
$begingroup$
OK, so this is saying that the first 2 elements will always have the same permutation?
$endgroup$
– Toby
Dec 13 '18 at 2:51
$begingroup$
@Toby Not quite .. it is saying that, given whatever the first two elements are, those two can either be put in the order where the first is smaller than the second, or the other way around, however the rest of the humbers are permuated after that. So, just as you have $1,2,4,3,5$, you also have $2,1,4,3,5$. Or, just as you have $3,5,1,2,4$, you have $5,3,1,2,4$. So they all come in pairs like at, where the sequence after the first two is exactly the same.
$endgroup$
– Bram28
Dec 13 '18 at 2:57
$begingroup$
@Toby Not quite .. it is saying that, given whatever the first two elements are, those two can either be put in the order where the first is smaller than the second, or the other way around, however the rest of the humbers are permuated after that. So, just as you have $1,2,4,3,5$, you also have $2,1,4,3,5$. Or, just as you have $3,5,1,2,4$, you have $5,3,1,2,4$. So they all come in pairs like at, where the sequence after the first two is exactly the same.
$endgroup$
– Bram28
Dec 13 '18 at 2:57
1
1
$begingroup$
@Toby Right, there are $(n-2)!$ ways to permuate the $n-2$ numbers after the first two. And yes, there are $n choose 2$ ways to pick the first two. But what I am saying is: it really doesn't matter what the exact number of permutations is or what the number of ways to pick the first two numbers is ... all you need to do is realize that whatever those numbers are with $a_1<a_2$, you get the same numbers for $a_2<a_1$ ... so each is half of all $n!$ permutations. No complicated calculations needed!
$endgroup$
– Bram28
Dec 13 '18 at 3:26
$begingroup$
@Toby Right, there are $(n-2)!$ ways to permuate the $n-2$ numbers after the first two. And yes, there are $n choose 2$ ways to pick the first two. But what I am saying is: it really doesn't matter what the exact number of permutations is or what the number of ways to pick the first two numbers is ... all you need to do is realize that whatever those numbers are with $a_1<a_2$, you get the same numbers for $a_2<a_1$ ... so each is half of all $n!$ permutations. No complicated calculations needed!
$endgroup$
– Bram28
Dec 13 '18 at 3:26
|
show 1 more comment
$begingroup$
The condition is only on the first 2, and not on the rest. Once the first two satisfy your requirement the reamining can be arranged in any way, that is $(n-2)!$ Now the first two have are two elements from $n$ with specific order (ascending), and no rearrangement which is ${nchoose 2}$. So it is the product of these two numbers, ${nchoose 2}times (n-2)!$ which is same as $frac12 n!$
answered Dec 13 '18 at 2:42
P VanchinathanP Vanchinathan
15.2k12136
$endgroup$
$begingroup$
Why is it nchoose2 for the first 2 elements?
$endgroup$
– Toby
Dec 13 '18 at 3:07
$begingroup$
This is the first lesson in counting. The number of ways of picking r elements (and not rearranging them) is n-choose-r. Not rearranging them can be guaranteed if we agree to stick to ascending order arrangement as the only way.
$endgroup$
– P Vanchinathan
Dec 13 '18 at 3:14
add a comment |
$begingroup$
The condition is only on the first 2, and not on the rest. Once the first two satisfy your requirement the reamining can be arranged in any way, that is $(n-2)!$ Now the first two have are two elements from $n$ with specific order (ascending), and no rearrangement which is ${nchoose 2}$. So it is the product of these two numbers, ${nchoose 2}times (n-2)!$ which is same as $frac12 n!$
answered Dec 13 '18 at 2:42
P VanchinathanP Vanchinathan
15.2k12136
$endgroup$
$begingroup$
Why is it nchoose2 for the first 2 elements?
$endgroup$
– Toby
Dec 13 '18 at 3:07
$begingroup$
This is the first lesson in counting. The number of ways of picking r elements (and not rearranging them) is n-choose-r. Not rearranging them can be guaranteed if we agree to stick to ascending order arrangement as the only way.
$endgroup$
– P Vanchinathan
Dec 13 '18 at 3:14
add a comment |
$begingroup$
The condition is only on the first 2, and not on the rest. Once the first two satisfy your requirement the reamining can be arranged in any way, that is $(n-2)!$ Now the first two have are two elements from $n$ with specific order (ascending), and no rearrangement which is ${nchoose 2}$. So it is the product of these two numbers, ${nchoose 2}times (n-2)!$ which is same as $frac12 n!$
answered Dec 13 '18 at 2:42
P VanchinathanP Vanchinathan
15.2k12136
$endgroup$
The condition is only on the first 2, and not on the rest. Once the first two satisfy your requirement the reamining can be arranged in any way, that is $(n-2)!$ Now the first two have are two elements from $n$ with specific order (ascending), and no rearrangement which is ${nchoose 2}$. So it is the product of these two numbers, ${nchoose 2}times (n-2)!$ which is same as $frac12 n!$
answered Dec 13 '18 at 2:42
P VanchinathanP Vanchinathan
15.2k12136
answered Dec 13 '18 at 2:42
P VanchinathanP Vanchinathan
15.2k12136
answered Dec 13 '18 at 2:42
P VanchinathanP Vanchinathan
15.2k12136
answered Dec 13 '18 at 2:42
P VanchinathanP Vanchinathan
15.2k12136
15.2k12136
$begingroup$
Why is it nchoose2 for the first 2 elements?
$endgroup$
– Toby
Dec 13 '18 at 3:07
$begingroup$
This is the first lesson in counting. The number of ways of picking r elements (and not rearranging them) is n-choose-r. Not rearranging them can be guaranteed if we agree to stick to ascending order arrangement as the only way.
$endgroup$
– P Vanchinathan
Dec 13 '18 at 3:14
add a comment |
$begingroup$
Why is it nchoose2 for the first 2 elements?
$endgroup$
– Toby
Dec 13 '18 at 3:07
$begingroup$
This is the first lesson in counting. The number of ways of picking r elements (and not rearranging them) is n-choose-r. Not rearranging them can be guaranteed if we agree to stick to ascending order arrangement as the only way.
$endgroup$
– P Vanchinathan
Dec 13 '18 at 3:14
$begingroup$
Why is it nchoose2 for the first 2 elements?
$endgroup$
– Toby
Dec 13 '18 at 3:07
$begingroup$
Why is it nchoose2 for the first 2 elements?
$endgroup$
– Toby
Dec 13 '18 at 3:07
$begingroup$
This is the first lesson in counting. The number of ways of picking r elements (and not rearranging them) is n-choose-r. Not rearranging them can be guaranteed if we agree to stick to ascending order arrangement as the only way.
$endgroup$
– P Vanchinathan
Dec 13 '18 at 3:14
$begingroup$
This is the first lesson in counting. The number of ways of picking r elements (and not rearranging them) is n-choose-r. Not rearranging them can be guaranteed if we agree to stick to ascending order arrangement as the only way.
$endgroup$
– P Vanchinathan
Dec 13 '18 at 3:14
add a comment |
$begingroup$
The number of all permutation is $n!$. Half of it with $a_1<a_2$ and the other half with $a_2<a_1$.
answered Dec 13 '18 at 2:36
user9077user9077
1,239612
$endgroup$
add a comment |
$begingroup$
The number of all permutation is $n!$. Half of it with $a_1<a_2$ and the other half with $a_2<a_1$.
answered Dec 13 '18 at 2:36
user9077user9077
1,239612
$endgroup$
add a comment |
$begingroup$
The number of all permutation is $n!$. Half of it with $a_1<a_2$ and the other half with $a_2<a_1$.
answered Dec 13 '18 at 2:36
user9077user9077
1,239612
$endgroup$
The number of all permutation is $n!$. Half of it with $a_1<a_2$ and the other half with $a_2<a_1$.
answered Dec 13 '18 at 2:36
user9077user9077
1,239612
answered Dec 13 '18 at 2:36
user9077user9077
1,239612
answered Dec 13 '18 at 2:36
user9077user9077
1,239612
answered Dec 13 '18 at 2:36
user9077user9077
1,239612
1,239612
add a comment |
add a comment |
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