Inviting 5 friends out of 11 to dinner, with restriction
I'm revising for finals and I have come across this following question:
a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.
b) Repeat a) but 2 of the friends are married and will not attend separately.
For a) I got ${11 choose 5} = frac{11!}{(11-5)! cdot 5!} = 462$.
I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then:
${9 choose 5} = frac{9!}{(9-5)! cdot 5!} = 126$, but I'm pretty sure this is wrong,
Could someone explain b) to me?
Thanks in advance!
probability combinatorics
edited Jun 14 '13 at 19:29
vadim123
75.6k897189
asked Apr 22 '12 at 14:54
Xabi
1713715
add a comment |
I'm revising for finals and I have come across this following question:
a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.
b) Repeat a) but 2 of the friends are married and will not attend separately.
For a) I got ${11 choose 5} = frac{11!}{(11-5)! cdot 5!} = 462$.
I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then:
${9 choose 5} = frac{9!}{(9-5)! cdot 5!} = 126$, but I'm pretty sure this is wrong,
Could someone explain b) to me?
Thanks in advance!
probability combinatorics
edited Jun 14 '13 at 19:29
vadim123
75.6k897189
asked Apr 22 '12 at 14:54
Xabi
1713715
add a comment |
I'm revising for finals and I have come across this following question:
a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.
b) Repeat a) but 2 of the friends are married and will not attend separately.
For a) I got ${11 choose 5} = frac{11!}{(11-5)! cdot 5!} = 462$.
I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then:
${9 choose 5} = frac{9!}{(9-5)! cdot 5!} = 126$, but I'm pretty sure this is wrong,
Could someone explain b) to me?
Thanks in advance!
probability combinatorics
edited Jun 14 '13 at 19:29
vadim123
75.6k897189
asked Apr 22 '12 at 14:54
Xabi
1713715
I'm revising for finals and I have come across this following question:
a) A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner.
b) Repeat a) but 2 of the friends are married and will not attend separately.
For a) I got ${11 choose 5} = frac{11!}{(11-5)! cdot 5!} = 462$.
I'm completely lost on b). I tried $n-2$ (minus two friends that are married) and then:
${9 choose 5} = frac{9!}{(9-5)! cdot 5!} = 126$, but I'm pretty sure this is wrong,
Could someone explain b) to me?
Thanks in advance!
probability combinatorics
probability combinatorics
edited Jun 14 '13 at 19:29
vadim123
75.6k897189
asked Apr 22 '12 at 14:54
Xabi
1713715
edited Jun 14 '13 at 19:29
vadim123
75.6k897189
asked Apr 22 '12 at 14:54
Xabi
1713715
edited Jun 14 '13 at 19:29
vadim123
75.6k897189
edited Jun 14 '13 at 19:29
vadim123
75.6k897189
edited Jun 14 '13 at 19:29
vadim123
75.6k897189
75.6k897189
asked Apr 22 '12 at 14:54
Xabi
1713715
asked Apr 22 '12 at 14:54
Xabi
1713715
asked Apr 22 '12 at 14:54
Xabi
1713715
1713715
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMM
9,11032848
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
Tomarinator
1,32111023
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
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Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMM
9,11032848
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMM
9,11032848
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMM
9,11032848
Hint: For (b), try subtracting from your answer to (a) all combinations of $5$ people that include exactly one of the two married friends.
answered Apr 22 '12 at 15:02
TMM
9,11032848
answered Apr 22 '12 at 15:02
TMM
9,11032848
answered Apr 22 '12 at 15:02
TMM
9,11032848
answered Apr 22 '12 at 15:02
TMM
9,11032848
9,11032848
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
10!(5!x5!)=252 462-252=210. I get it Thanks! :)
– Xabi
Apr 22 '12 at 15:07
add a comment |
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
Tomarinator
1,32111023
add a comment |
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
Tomarinator
1,32111023
add a comment |
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
Tomarinator
1,32111023
for part b,
the woman has 2 choices , either to invite the couple or not invite them ,
CASE1: she invites them
then she has to choose 3 friends out of 9 (2 out of 11 are already chosen), which she can do in ${9 choose 3}$ ways
CASE2: when she doesnt invites the couple
then she has to choose 5 friends out of 9(2out of 11 are not to be chosen), which she can do in ${9 choose 5}$ ways
answered Apr 22 '12 at 15:06
Tomarinator
1,32111023
answered Apr 22 '12 at 15:06
Tomarinator
1,32111023
answered Apr 22 '12 at 15:06
Tomarinator
1,32111023
answered Apr 22 '12 at 15:06
Tomarinator
1,32111023
1,32111023
add a comment |
add a comment |
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