2
$\begingroup$

I'm having trouble figuring out the permutations and how to properly multiply them together for this problem.

A computer programming team has $13$ members.

a.) How many ways can a group of seven be chosen to work on a project?

b.) Suppose seven team members are women and six are men. i) How many groups of seven can be chosen that contain four women and three men?

b. is where I'm having the most trouble, I think a. is just $P(13,7)$ On b. I know its suppose to be written something like ${7 \choose 4}{7 \choose 3}$

Then how do you multiply the two sets is what I'm wondering? Book didn't help much and can't find a clear answer online.

Later it asks "How many groups of seven can be chosen that contain at least one man?" and "How many groups of seven can be chosen that contain at most three women?"

Thanks for the help - sorry for all the writing.

$\endgroup$
  • $\begingroup$ Please take some time to read the formatting page to help you with writing your questions clearly. $\endgroup$ – Mnifldz Aug 8 '15 at 17:54
  • $\begingroup$ Hint for b.): Think of the $\texttt{hypergeometric distribution}$ But do not divide by $N \choose n$ $\endgroup$ – callculus Aug 8 '15 at 17:57
  • $\begingroup$ Figured out how to write the two line notation. Edited the question. $\endgroup$ – androidguy Aug 8 '15 at 18:00
  • $\begingroup$ So is it $N=7$ and $n=4$? What would $K$ and $k$ be? $\endgroup$ – androidguy Aug 8 '15 at 18:05
  • $\begingroup$ @androidguy $N=13, K=6, n=7, k=3$ and Therefore $N-K=7$, $n-k=4$ $\endgroup$ – callculus Aug 8 '15 at 18:29
4
$\begingroup$

First it might be helpful to remember that the binomial coefficients are given by

$$ {n \choose k} \;\; =\;\; \frac{n!}{k!(n-k)!}. $$

Assuming that in your notation $P(13,7) = {13 \choose 7}$ then you're correct for the first part of the question. For the second part, you need to pick $4$ women and $3$ men. You have the right idea in what (I believe) you're writing, but it needs to be refined a bit.

Note that we need to pick $4$ out of seven women and there are ${7\choose 4}$ ways to do this. Similarly there are ${6\choose 3}$ ways to choose the men. Since the choice of women can be done independently of the choice of men (every potential group of women can be combined with any of the possible groups of men) we simply multiply these quantities together. The answer should be ${7\choose 4}{6\choose 3}$.

When we get to questions of there being "at most" or "at least" a certain number of men and/or women, we need to consider each case separately and then add all of the possible numbers together. To illustrate, consider the problem of computing the number of groups containing at least one man. We can make a list of all of the different possible numbers simply by considering each case:

  • The number of groups with zero men is ${7 \choose 7}{6\choose 0} = 1$, since all $7$ women just get chosen.
  • The number of groups with one man is ${7\choose 6}{6\choose 1} = 42$.
  • The number of groups with two men is ${7\choose 5}{6\choose 2}$.
  • etc.

To find the number of cases with at least one man, we add up the numbers of groups with one man, two men, three men, all the way to six men. Another way of computing this number too is to realize that the total number of possible groups is just ${13\choose 7}$. The only possible case that we are eliminating from our analysis is when all $7$ women get chosen, thus we can equivalently come to the conclusion that the number of groups with at least one man is ${13\choose 7} - {7\choose 7}{6\choose 0}$.

$\endgroup$
  • $\begingroup$ Thank you so much, very thorough answer. Really appreciate it! $\endgroup$ – androidguy Aug 8 '15 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.