did i solve these counting problems correctly? [closed]
$begingroup$
nine people (Ann, Ben, Cal, Dot, Ed, Fran, Gail, Hal, and Ida) are in a room. Five of them stand in a row for a picture.
a) In how many ways can this be done if both Ed and Gail are in the picture?
P(5,2)= 5×4 = 20
b) In how many ways can this be done if neither Ed nor Fran are in the picture?
P(5,3)= 5×4×3= 60
c) In how many ways can this be done if Dot is on the left end and Ed is on the
right end?
P(5,3)= 5×4×3= 60
d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?
P(5,4) + P(5,4)= 120+120= 240
e) In how many ways can this be done if Ed and Gail are in the picture, standing next to each other?
P(5,4)= 120
Thank you
discrete-mathematics
asked Nov 28 '18 at 13:03
Rose .MRose .M
83
$endgroup$
closed as too broad by amWhy, choco_addicted, Adrian Keister, user10354138, José Carlos Santos Nov 28 '18 at 15:43
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
nine people (Ann, Ben, Cal, Dot, Ed, Fran, Gail, Hal, and Ida) are in a room. Five of them stand in a row for a picture.
a) In how many ways can this be done if both Ed and Gail are in the picture?
P(5,2)= 5×4 = 20
b) In how many ways can this be done if neither Ed nor Fran are in the picture?
P(5,3)= 5×4×3= 60
c) In how many ways can this be done if Dot is on the left end and Ed is on the
right end?
P(5,3)= 5×4×3= 60
d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?
P(5,4) + P(5,4)= 120+120= 240
e) In how many ways can this be done if Ed and Gail are in the picture, standing next to each other?
P(5,4)= 120
Thank you
discrete-mathematics
asked Nov 28 '18 at 13:03
Rose .MRose .M
83
$endgroup$
closed as too broad by amWhy, choco_addicted, Adrian Keister, user10354138, José Carlos Santos Nov 28 '18 at 15:43
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to Math StackExchange. Usually you will get better help if you show some effort and/or focus more specifically on what you are having trouble with. You are less likely to get help if you post what appears to be verbatim homework with no show of effort.
$endgroup$
– paw88789
Nov 28 '18 at 13:12
add a comment |
$begingroup$
nine people (Ann, Ben, Cal, Dot, Ed, Fran, Gail, Hal, and Ida) are in a room. Five of them stand in a row for a picture.
a) In how many ways can this be done if both Ed and Gail are in the picture?
P(5,2)= 5×4 = 20
b) In how many ways can this be done if neither Ed nor Fran are in the picture?
P(5,3)= 5×4×3= 60
c) In how many ways can this be done if Dot is on the left end and Ed is on the
right end?
P(5,3)= 5×4×3= 60
d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?
P(5,4) + P(5,4)= 120+120= 240
e) In how many ways can this be done if Ed and Gail are in the picture, standing next to each other?
P(5,4)= 120
Thank you
discrete-mathematics
asked Nov 28 '18 at 13:03
Rose .MRose .M
83
$endgroup$
nine people (Ann, Ben, Cal, Dot, Ed, Fran, Gail, Hal, and Ida) are in a room. Five of them stand in a row for a picture.
a) In how many ways can this be done if both Ed and Gail are in the picture?
P(5,2)= 5×4 = 20
b) In how many ways can this be done if neither Ed nor Fran are in the picture?
P(5,3)= 5×4×3= 60
c) In how many ways can this be done if Dot is on the left end and Ed is on the
right end?
P(5,3)= 5×4×3= 60
d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?
P(5,4) + P(5,4)= 120+120= 240
e) In how many ways can this be done if Ed and Gail are in the picture, standing next to each other?
P(5,4)= 120
Thank you
discrete-mathematics
discrete-mathematics
asked Nov 28 '18 at 13:03
Rose .MRose .M
83
asked Nov 28 '18 at 13:03
Rose .MRose .M
83
edited Nov 28 '18 at 13:15
Rose .M
asked Nov 28 '18 at 13:03
Rose .MRose .M
83
asked Nov 28 '18 at 13:03
Rose .MRose .M
83
asked Nov 28 '18 at 13:03
Rose .MRose .M
83
83
closed as too broad by amWhy, choco_addicted, Adrian Keister, user10354138, José Carlos Santos Nov 28 '18 at 15:43
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by amWhy, choco_addicted, Adrian Keister, user10354138, José Carlos Santos Nov 28 '18 at 15:43
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Welcome to Math StackExchange. Usually you will get better help if you show some effort and/or focus more specifically on what you are having trouble with. You are less likely to get help if you post what appears to be verbatim homework with no show of effort.
$endgroup$
– paw88789
Nov 28 '18 at 13:12
add a comment |
$begingroup$
Welcome to Math StackExchange. Usually you will get better help if you show some effort and/or focus more specifically on what you are having trouble with. You are less likely to get help if you post what appears to be verbatim homework with no show of effort.
$endgroup$
– paw88789
Nov 28 '18 at 13:12
$begingroup$
Welcome to Math StackExchange. Usually you will get better help if you show some effort and/or focus more specifically on what you are having trouble with. You are less likely to get help if you post what appears to be verbatim homework with no show of effort.
$endgroup$
– paw88789
Nov 28 '18 at 13:12
$begingroup$
Welcome to Math StackExchange. Usually you will get better help if you show some effort and/or focus more specifically on what you are having trouble with. You are less likely to get help if you post what appears to be verbatim homework with no show of effort.
$endgroup$
– paw88789
Nov 28 '18 at 13:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I post the answer with explanation for each section below. Hope it will be clear enough.
a) In how many ways can this be done if both Ed and Gail are in the picture?
Given that Ed and Gail are already in picture, you need 3 more from the rest of the people in ${7 choose 3}$ =35 ways. Also, 5 people can stand in a picture in $5!$ ways. So the answer is $35*120=4200$.
b) In how many ways can this be done if neither Ed nor Fran are in the picture?
Since, you need to choose 5 from 7 people, you can do it in ${7 choose 5} = 42$ ways. Also, since ordering matters, you need to multiply this value by $5!=120$. Answer: $42*120=5040$.
c) In how many ways can this be done if Dot is on the left end and Ed is on the right end?
Now you need to feel 3 spots in the middle of picture from 7 people that are not already in picture. That is ${7 choose 3}=35$. While seats on the left and right are reserved, you can still arrange the people in the middle in $3!=6$ ways. Hence, asnwer: $35*6=210$.
d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?
We can calculate in different ways, but the way I do is to calculate the case Hal is in the picture but not Ida, and Ida is in the picture but not Hal cases separately, and add them up.
Hal is in picture, but not Ida: ${7 choose 4}=35$ ways of choosing people, and $5!$ ordering. Answer: $35*120=4200$
Ida is in picture, but not Hal: Exactly the same as above with the names flipped. Hence, answer: 4200.
Overall: 8400
In how many ways can this be done if Ed and Gail are in the picture, standing next to each other?
Similarly, here you need to choose 3 people from the rest (${7 choose 3}=35$. But this time, we will treat Ed and Gail as a single body, and the number of different orderings for 4 people is $4!$. However, since, Ed and Gail can exchange their seats with each other, we still need to multiple overall result to 2. Hence, answer: $35*24*2=960$.
answered Nov 28 '18 at 13:54
Vorashil FarzaliyevVorashil Farzaliyev
362
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I post the answer with explanation for each section below. Hope it will be clear enough.
a) In how many ways can this be done if both Ed and Gail are in the picture?
Given that Ed and Gail are already in picture, you need 3 more from the rest of the people in ${7 choose 3}$ =35 ways. Also, 5 people can stand in a picture in $5!$ ways. So the answer is $35*120=4200$.
b) In how many ways can this be done if neither Ed nor Fran are in the picture?
Since, you need to choose 5 from 7 people, you can do it in ${7 choose 5} = 42$ ways. Also, since ordering matters, you need to multiply this value by $5!=120$. Answer: $42*120=5040$.
c) In how many ways can this be done if Dot is on the left end and Ed is on the right end?
Now you need to feel 3 spots in the middle of picture from 7 people that are not already in picture. That is ${7 choose 3}=35$. While seats on the left and right are reserved, you can still arrange the people in the middle in $3!=6$ ways. Hence, asnwer: $35*6=210$.
d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?
We can calculate in different ways, but the way I do is to calculate the case Hal is in the picture but not Ida, and Ida is in the picture but not Hal cases separately, and add them up.
Hal is in picture, but not Ida: ${7 choose 4}=35$ ways of choosing people, and $5!$ ordering. Answer: $35*120=4200$
Ida is in picture, but not Hal: Exactly the same as above with the names flipped. Hence, answer: 4200.
Overall: 8400
In how many ways can this be done if Ed and Gail are in the picture, standing next to each other?
Similarly, here you need to choose 3 people from the rest (${7 choose 3}=35$. But this time, we will treat Ed and Gail as a single body, and the number of different orderings for 4 people is $4!$. However, since, Ed and Gail can exchange their seats with each other, we still need to multiple overall result to 2. Hence, answer: $35*24*2=960$.
answered Nov 28 '18 at 13:54
Vorashil FarzaliyevVorashil Farzaliyev
362
$endgroup$
add a comment |
$begingroup$
I post the answer with explanation for each section below. Hope it will be clear enough.
a) In how many ways can this be done if both Ed and Gail are in the picture?
Given that Ed and Gail are already in picture, you need 3 more from the rest of the people in ${7 choose 3}$ =35 ways. Also, 5 people can stand in a picture in $5!$ ways. So the answer is $35*120=4200$.
b) In how many ways can this be done if neither Ed nor Fran are in the picture?
Since, you need to choose 5 from 7 people, you can do it in ${7 choose 5} = 42$ ways. Also, since ordering matters, you need to multiply this value by $5!=120$. Answer: $42*120=5040$.
c) In how many ways can this be done if Dot is on the left end and Ed is on the right end?
Now you need to feel 3 spots in the middle of picture from 7 people that are not already in picture. That is ${7 choose 3}=35$. While seats on the left and right are reserved, you can still arrange the people in the middle in $3!=6$ ways. Hence, asnwer: $35*6=210$.
d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?
We can calculate in different ways, but the way I do is to calculate the case Hal is in the picture but not Ida, and Ida is in the picture but not Hal cases separately, and add them up.
Hal is in picture, but not Ida: ${7 choose 4}=35$ ways of choosing people, and $5!$ ordering. Answer: $35*120=4200$
Ida is in picture, but not Hal: Exactly the same as above with the names flipped. Hence, answer: 4200.
Overall: 8400
In how many ways can this be done if Ed and Gail are in the picture, standing next to each other?
Similarly, here you need to choose 3 people from the rest (${7 choose 3}=35$. But this time, we will treat Ed and Gail as a single body, and the number of different orderings for 4 people is $4!$. However, since, Ed and Gail can exchange their seats with each other, we still need to multiple overall result to 2. Hence, answer: $35*24*2=960$.
answered Nov 28 '18 at 13:54
Vorashil FarzaliyevVorashil Farzaliyev
362
$endgroup$
add a comment |
$begingroup$
I post the answer with explanation for each section below. Hope it will be clear enough.
a) In how many ways can this be done if both Ed and Gail are in the picture?
Given that Ed and Gail are already in picture, you need 3 more from the rest of the people in ${7 choose 3}$ =35 ways. Also, 5 people can stand in a picture in $5!$ ways. So the answer is $35*120=4200$.
b) In how many ways can this be done if neither Ed nor Fran are in the picture?
Since, you need to choose 5 from 7 people, you can do it in ${7 choose 5} = 42$ ways. Also, since ordering matters, you need to multiply this value by $5!=120$. Answer: $42*120=5040$.
c) In how many ways can this be done if Dot is on the left end and Ed is on the right end?
Now you need to feel 3 spots in the middle of picture from 7 people that are not already in picture. That is ${7 choose 3}=35$. While seats on the left and right are reserved, you can still arrange the people in the middle in $3!=6$ ways. Hence, asnwer: $35*6=210$.
d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?
We can calculate in different ways, but the way I do is to calculate the case Hal is in the picture but not Ida, and Ida is in the picture but not Hal cases separately, and add them up.
Hal is in picture, but not Ida: ${7 choose 4}=35$ ways of choosing people, and $5!$ ordering. Answer: $35*120=4200$
Ida is in picture, but not Hal: Exactly the same as above with the names flipped. Hence, answer: 4200.
Overall: 8400
In how many ways can this be done if Ed and Gail are in the picture, standing next to each other?
Similarly, here you need to choose 3 people from the rest (${7 choose 3}=35$. But this time, we will treat Ed and Gail as a single body, and the number of different orderings for 4 people is $4!$. However, since, Ed and Gail can exchange their seats with each other, we still need to multiple overall result to 2. Hence, answer: $35*24*2=960$.
answered Nov 28 '18 at 13:54
Vorashil FarzaliyevVorashil Farzaliyev
362
$endgroup$
I post the answer with explanation for each section below. Hope it will be clear enough.
a) In how many ways can this be done if both Ed and Gail are in the picture?
Given that Ed and Gail are already in picture, you need 3 more from the rest of the people in ${7 choose 3}$ =35 ways. Also, 5 people can stand in a picture in $5!$ ways. So the answer is $35*120=4200$.
b) In how many ways can this be done if neither Ed nor Fran are in the picture?
Since, you need to choose 5 from 7 people, you can do it in ${7 choose 5} = 42$ ways. Also, since ordering matters, you need to multiply this value by $5!=120$. Answer: $42*120=5040$.
c) In how many ways can this be done if Dot is on the left end and Ed is on the right end?
Now you need to feel 3 spots in the middle of picture from 7 people that are not already in picture. That is ${7 choose 3}=35$. While seats on the left and right are reserved, you can still arrange the people in the middle in $3!=6$ ways. Hence, asnwer: $35*6=210$.
d) In how many ways can this be done if Hal or Ida (but not both) are in the picture?
We can calculate in different ways, but the way I do is to calculate the case Hal is in the picture but not Ida, and Ida is in the picture but not Hal cases separately, and add them up.
Hal is in picture, but not Ida: ${7 choose 4}=35$ ways of choosing people, and $5!$ ordering. Answer: $35*120=4200$
Ida is in picture, but not Hal: Exactly the same as above with the names flipped. Hence, answer: 4200.
Overall: 8400
In how many ways can this be done if Ed and Gail are in the picture, standing next to each other?
Similarly, here you need to choose 3 people from the rest (${7 choose 3}=35$. But this time, we will treat Ed and Gail as a single body, and the number of different orderings for 4 people is $4!$. However, since, Ed and Gail can exchange their seats with each other, we still need to multiple overall result to 2. Hence, answer: $35*24*2=960$.
answered Nov 28 '18 at 13:54
Vorashil FarzaliyevVorashil Farzaliyev
362
answered Nov 28 '18 at 13:54
Vorashil FarzaliyevVorashil Farzaliyev
362
answered Nov 28 '18 at 13:54
Vorashil FarzaliyevVorashil Farzaliyev
362
answered Nov 28 '18 at 13:54
Vorashil FarzaliyevVorashil Farzaliyev
362
362
add a comment |
add a comment |
$begingroup$
Welcome to Math StackExchange. Usually you will get better help if you show some effort and/or focus more specifically on what you are having trouble with. You are less likely to get help if you post what appears to be verbatim homework with no show of effort.
$endgroup$
– paw88789
Nov 28 '18 at 13:12