Combinations and permutations when separating into groups

Question

1. If I have 30 people and I want to form them into 3 groups. One of size 10, one of size 5 and one of size 15. How many ways can I do this?

2. Similarly what if I have 15 boys and 15 girls and the 10 person group has to have 5 of each?

For part 1, I thought I would do 30C10 + 20C5 + 15C15 and that it would equal if I arranged it any other way, like 30C5 + 25C10 + 15C15, but I get different answers! Help?

Answer 1

One way to think of the first part is to have the people numbered 1 to 30, and they do not move, now count the number of arrangements of labels which assign them into one of the three groups.

111111111122222333333333333333

this is arranging 30 labels with repetitions of 10, 5, and 15 so the total number off arrangements (and the total number of assignments of the 30 people into the 3 groups) is given by 30!/(10!5!15!)

Answer 2

1. If I have 30 people and I want to form them into 3 groups. One of size 10, one of size 5 and one of size 15. How many ways can I do this?
2. Similarly what if I have 15 boys and 15 girls and the 10 person group has to have 5 of each?

1.

It's a straight multiset selection: $${30 \choose 15, 10, 5} = {30\choose 15}{15\choose 10}{5\choose 5} = {^{30}C_{15}}\cdot{^{15}C_{10}}\cdot{^{5}C_{5}} = \frac{30!}{15!\,10!\,5!}$$ This counts the ways to select a teams of 15 from 30, times the ways to select a team of 10 from the remaining 15, and times the ways* to select at team of 5 from the remaining 5.

2.

Choose the boys and girls for the balanced team; then choose the two other teams from the remaining students.

$${15\choose 5}^2{20\choose 15}{5\choose 5} = {^{15}C_{5}}^2\cdot{^{20}C_{15}}\cdot{^{5}C_{5}} = \frac{15!\, 20!}{5!^3\, 10!^2}$$

This counts the ways to select a half-team of 5 boys from 15 boys, times the ways to select the other half-team of 5 girls from 15 girls, times the ways to select a team of 10 from the remaining 15 people, and times the ways* to select at team of 5 from the remaining 5 people.

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(Note: * Counting the ways to select 5 from 5 is included for completeness, though it is obviously just 1.)