# Simple (probably) math question

**URL:** <https://boards.straightdope.com/t/simple-probably-math-question/258625>\
**Category:** Factual Questions\
**Created:** [August 6, 2004, 3:00pm UTC](https://boards.straightdope.com/t/simple-probably-math-question/258625 "2004-08-06T15:00:36Z")\
**Posts on this page:** 3\
**Page:** 1

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**Author:** ![Thrillhouse15](https://avatars.discourse-cdn.com/v4/letter/t/ad7895/32.png) [@Thrillhouse15](https://boards.straightdope.com/u/Thrillhouse15)\
**Post date:** [August 6, 2004, 3:00pm UTC](https://boards.straightdope.com/t/simple-probably-math-question/258625/1 "2004-08-06T15:00:36Z")

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I’m sure there’s an easy way to do this, but I can’t seem to remember how. Maybe some math doper can help me out.

I have a class of 14 boys and 8 girls. We’re going on a field trip, and I want to put on one of the buses 10 boys and 5 girls (the rest of the kids ride on the other bus- doesn’t matter why). The order in which these 10 boys and 5 girls sit doesn’t matter, there just has to be 10 boys and 5 girls. I want to know how many possible different combinations of the 10 boys and 5 girls I could have on that bus. How do I figure this out?

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**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\
**Post date:** [August 6, 2004, 3:15pm UTC](https://boards.straightdope.com/t/simple-probably-math-question/258625/2 "2004-08-06T15:15:22Z")

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You coulnt to number of different ways of choosing 10 out of 14, and the number of different ways of choosing 5 out of 8, the multiply them together:

(14 x 13 x 12 x 11) / (2 x 3 x 4)  
times  
(8 x 7 x 6) / (2 x 3)

i.e. 1001 times 56

i.e. 56,056

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**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\
**Post date:** [August 6, 2004, 5:42pm UTC](https://boards.straightdope.com/t/simple-probably-math-question/258625/3 "2004-08-06T17:42:20Z")

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**Giles** is correct, and the following page (see Combinations Rule) explains the derivation of how many ways to select r items from a population of N. It gives you the general solution for this kind of problem. In case you actually wanted it.

[http://www.cquest.utoronto.ca/geog/ggr270y/notes/not08a.html](http://www.cquest.utoronto.ca/geog/ggr270y/notes/not08a.html)

And once you have the combinations of boys and combinations of girls, you simply multiply them to get all combinations of the combinations, as Giles states.