# Combinations math problem

1. May 4, 2014

### joshmccraney

1. The problem statement, all variables and given/known data
there are 10 women and 12 men. 5 pairs are to be made, each pair having one man and one women. how many pairings are possible.

3. The attempt at a solution
first let's choose men and women from each set. there are ${10 \choose 5}$ women and ${12 \choose 5}$ men. now that we have these, what should i be thinking? i know the answer is ${12 \choose 5} \times {10 \choose 5} \times 5!$. why multiply by the $5!$?

thanks!

2. May 4, 2014

### LCKurtz

Your binomial coefficients give the number of ways of choosing 5 women and 5 men. Now you have to figure out how many different pairs you can make with them.

3. May 4, 2014

### xiavatar

In other words once you have chose the five men and five woman, how many ways can you order them man,woman,man,woman, etcetera...

4. May 5, 2014

### joshmccraney

how many different pairs...this seems tough. well, i'm thinking of 5 positions. first, for the men there are $5!$ different ways to seat them. now for the women, there are also $5!$ ways to sit them. i think the answer is now $5!+5!$. however, since women/man is the same as man/women, we divide by $2!$?

is this correct?

thanks!

5. May 5, 2014

### LCKurtz

No. Normally counting like that you would multiply the two numbers. But that isn't the correct analysis for this problem anyway. Look at it this way. All the men are seated and it doesn't matter what order. Now how many ways can you seat the ladies on their laps? There are your pairs.

6. May 5, 2014

### joshmccraney

Alright this is great! Makes perfect sense! Thanks!!!