There are 32 women and 26 men in a dance course. They form pairs using genders (1 male 1 female in each pair). How many different pairs you

I am not terribly good with prob but I will give it a shot.

chose a man any man will do so the prob is one.  He can have any 32 partners.

Grab a 2nd man.  He can have any of 31 partners 

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Grab the last man and he can have any of 7 partners

There are 6 women left over

So perhaps there are 32!/6!   ways.

$${\frac{{\mathtt{32}}{!}}{{\mathtt{6}}{!}}} = {\mathtt{365\,459\,495\,741\,241\,014\,121\,136\,128\,000\,000}}$$

boy that is a lot of possible pairs.   

Anyone want to comment  - or shoot me down in flames?

Hello, I should've told earlier, but couldn't edit the post. The answer is (according to the book) 507. However I'm wondering how to calculate it?

And thanks for trying Melody, but it doesn't match the book's answer

We need more answers here.    !!   Beware it is  Probability !!

Notice that the first man has 32 choices, and the second man has 31 choices, and the 3rd man has 30 choices, etc.  By the time we get to the 26th man, he has 7 choices, thusly...

32  + 31 + 30+....+ 7 

1        2      3          26 

So, the number of choices  just equals the sum of the digits from 7 to 32....or, put another way...the sum of the first 32 digits  less the sum of the first 6 digits  =

32(33) / 2  -  6(7)/ 2  =

528            -   21   =

507

I think that's it.... [ I hope !! ]

Yes that makes sense Chris.  It is very unusual to use addition rather than multiplication.

But it does make sense in this instance.

Can you think what question I may have answered?

Actually, Melody...the answer seems like it should be WAY more than 507....I just kind of "played around" with it......

I think your answer approaches the number of hours that I have gone without sleep....LOL!!!

Maybe it is the sum of how many hours we have each gone without sleep.  :)