Combination problem (with exclude candidate)

tony24810

In how many ways can a committee of five be chosen from 11 people, if two particular people will not work on the same committee (that is, if one is included, the other must be excluded).

I have attempted this questions with differents approaches and yielded the same answers, but my answer disagree with the answer from the book.

I would really appreciate if someone will point out if there're flaws in my arguments or if the book is wrong.


method 1:

suppose the group of 11 is split into 3 groups: guy1, guy2 and the rest(9 persons)

number of ways to form committe with guy1 = 1 * 9C4
number of ways to form committe with guy2 = 1 * 9C4
number of ways to form committe from 'the rest' = 9C5

number of ways = 9C4 + 9C4 + 9C5 = 126+126+126=378

method 2:

all combinations - the combinations that contain both guy1 and guy2
= 11C5 - 9C3
= 462 - 84
=378



The answer from the book is 252.

Please someone help! thanks!











the next question is similar:

In how many ways may a team of four oarsmen be chosen from a panel of 13 people if two individuals refuse to work on the same rowing team as each other?

my solution = ways with guy1 + ways with guy2 + ways without those 2
= 11C3 + 11C3 + 11C4
= 165 + 165 + 330
= 660


but the book says 495

haruspex

The book is definitely wrong. In each case, you can make what is obviously an underestimate by simply eliminating one of the protagonists, leaving 10C5 in the first and 12C4 in the second. Intriguingly, these produce the book answers.

mfb

I can confirm your numbers, and have no idea how the book got its numbers.

Edit: I really checked this for 6 minutes?

tony24810

Thanks a lot! I was starting to wonder if I should re do a statistic course lol.

Robert1986

Perhaps whoever answered the question to the book assumed that one of the two people must be on the committee.

mfb

This gives the book's answer of 252 for the first problem, but the answer to the second problem would be 330 in this case (2*(11 choose 3)).

You get the numbers from the book if you ignore every combination with guy 2:
252=(10 choose 5)
495= (12 choose 4)

Or, similar, if you consider (guy1, n others) and (guy2, n others) as one case (why??).