# Thread: Simple problem combinations permutations...

1. ## Simple problem combinations permutations...

Here's a problem I have to solve. I found what I think is the solution, but wanted to see how someone comfortable with math would solve it. It's fairly simple.

There are 7 women and 9 men.
a. How would you choose a 5 member group where at least one of those members is a woman?

b. How would you choose a 5 member group where at least one woman and one man had to be in the group?

This was from a lesson on r-combinations and r-permutations.

2. Originally Posted by webhound2
Here's a problem I have to solve. I found what I think is the solution, but wanted to see how someone comfortable with math would solve it. It's fairly simple.

There are 7 women and 9 men.
a. How would you choose a 5 member group where at least one of those members is a woman?

b. How would you choose a 5 member group where at least one woman and one man had to be in the group?

This was from a lesson on r-combinations and r-permutations.

As you already solved it show your work and then we'll be able to comment on it.

Tonio

3. I should clarify that the exact phrasing of the questions is "how many ways are there to choose 5 members"

I thought there was 7 ways to choose 1 woman, which left four choices to be made out of a pool of 15 persons. So I came up with 7 X 15 X 14 X 13 X 12 = 229320.

For b I did something similar and came up with 7 X 9 X 14 X 13 X 12 = 137592.

I attempted using the formula $\displaystyle \frac{n!}{r!(n!-r!)}$ but the answers I cam eup with didn't make sense to me. Though I believe that the formula is only useful when you don't care about the order, and in this case it seems that order matters.

4. Originally Posted by webhound2
I thought there was 7 ways to choose 1 woman, which left four choices to be made out of a pool of 15 persons. So I came up with 7 X 15 X 14 X 13 X 12 = 229320.
Trouble is, many of those supposedly $\displaystyle \binom{15}{4}$ different ways are really the same. So you need to tabulate:

$\displaystyle \begin{tabular}{|c|c|c|c|} \hline no. of W & no. of ways & no. of M & no. of ways \\ \hline 5 &$\binom{7}{5}$& 0 &$\binom{9}{0}$\\ \hline 4 & & 1 & \\ \hline 3 & & 2 & \\ \hline 2 & & 3 & \\ \hline 1 & & 4 & \\ \hline \end{tabular}$

Originally Posted by webhound2
$\displaystyle \frac{n!}{r!(n!-r!)}$
should be

$\displaystyle \frac{n!}{r!(n-r)!}$

5. Hello, webhound2!

There are 7 women and 9 men.
a. How many ways can we choose a 5-member group with at least one woman?

There are: .$\displaystyle \displaystyle _{16}C_5 \:=\:{16\choose5} \:=\:\frac{16!}{5!\,11!} \:=\:4368 \text{ possible groups.}$

There are: .$\displaystyle \displaystyle _9C_5 \:=\:{9\choose5} \:=\:\frac{9!}{5!\,4!} \:=\:126\text{ groups with }no\text{ women (all men).}$

Therefore, there are: .$\displaystyle 4368 - 126 \:=\:4242\text{ groups with at least one woman.}$

b. How many ways can we choose a 5-member group
with at least one woman and one man?

There are: .$\displaystyle 126\text{ groups which are all men.}$

There are: .$\displaystyle \displaystyle _7C_5 \:=\:\frac{7!}{5!\,2!} \:=\:21\text{ groups whch are all women.}$

Hence, there are: .$\displaystyle 126 + 21 \,=\,147\text{ single-gender groups.}$

Therefore, there are: .$\displaystyle 4368 - 147 \:=\:4221\text{ coed groups.}$

6. @Soroban - since the wording is "how many different ways are there to select a group of 5" imply that order matters here and we are looking for permutations and not combinations. My understanding is that if it had said "how many different groups can be chosen, then you approach would be the way to go.

Still, that's cool way to find it that I hadn't considered before.

@tom - I don't know how to tabulate. It isn't something we've covered.

7. That reminds me of the first time I was asked to 'tabulate' and it really scared me! It only means draw up a table!

Soroban kind of made a table, too, but he noticed that you only need two rows (one for no restriction and one for men only - to be subtracted. Which is much simpler and better). So he didn't bother with borders and stuff.

I very much doubt that you're interpretation is right - it would have been specified that the members should be caring where they sat (say).

8. That's a good point, I think I understand. If they were looking for permutations, would my approach work?

And thank you both, for you help.

9. Originally Posted by webhound2
That's a good point, I think I understand. If they were looking for permutations, would my approach work?
No, you'd still want to follow Soroban's method but of course use

$\displaystyle \frac{n!}{(n-r)!}$

instead of

$\displaystyle \frac{n!}{r!(n-r)!}$

Edit: to see the problem with your method, think it through...

You choose any one of the seven women, as you say. Then there are however many ways of choosing the rest from the remaining 15 people. Fine. But then you choose a different woman to be the first to be chosen in your next (of seven) similar rounds of choosing. But in this round you're going to end up choosing many groups that were already counted in the first round. (And if you care about seating you'll end up counting already-counted seating plans.)

10. Originally Posted by webhound2
@Soroban - since the wording is "how many different ways are there to select a group of 5" imply that order matters here and we are looking for permutations and not combinations.
That is simply not true. That is standard mathematical wording. Ask any author of mathematics text material on counting

There a general guide for writing mathematics contest questions. At least in the North America that is followed by most publishers and editors. I have reviewed too many textbooks for publishers not to know that.

If the author of the question means for permutations to be used the the question should have read "How many ways can a line (queue) of five be formed from seven people".

I am not saying that your author did not mean for you to use permutations.
What I am saying is that Soroban answered your question in the same way that anyone who has been active in this area would have.