Question

Suppose we have a group of 10 men and 15 women.

(a)[4 Pts]How many ways can we line them up so that no two men are next to eachother?Hint: Line up the women first and then place the men between them.

(b)[4 Pts]How many ways can a committee of eight be chosen which contains atleast three men and three women?

(c)[4 Pts]How many ways can we pick a president and secretary, a social committeeof four, a sports committee of five, and three identical working groups of size three?

Answer 1

a) 15!×16×15×...×7

b)
`\binom{15}{3}\binom{10}{3}\binom{19}{2}`.

c)
`25\cdot 24\cdot \binom{23}{4}\binom{19}{5}\binom{14}{3}\binom{11}{3}\binom{9}{3}`

Explanation:

a) There are 15! ways to line the women first. Now, a man can be placed either to the left of some woman (giving 15 choices) or to the right of the last woman on the line (1 additional choice). So the number of ways of placing the men is 16×15×...×7, that is, 10-permutations of 16. Then the number of ways to line everyone is 15!×16×15×...×7.

b) First, choose 3 women of said committee. There are 15 women and we must choose 3, then there are
`\binom{15}{3}` ways of doing this. Afterwards, choose 3 men for the commitee. There are 10 men, so the number of ways of doing this is
`\binom{10}{3}`. The commitee must have 8 people and we already chose 6, so we must choose 2 more persons. There are 25 people in total and we chose 6 of them before, so there are 19 who aren't in the committee. Then the number of ways of completing the commitee is
`\binom{19}{2}`. The number of ways of doing all of this (and creating the commitee) is
`\binom{15}{3}\binom{10}{3}\binom{19}{2}`.

c) There are 25 choices for a president. After choosing a president, there are 24 choices for the secretary. 23 people remain, then there are
`\binom{23}{4}` ways of choosing the social commitee. We are left with 19 people, thus there are
`\binom{19}{5}` ways to form the social commitee. Similarly, because 14 people haven't been selected, there are
`\binom{14}{3}\binom{11}{3}\binom{9}{3}` ways to form the working groups. We conclude that there are
`25\cdot 24\cdot \binom{23}{4}\binom{19}{5}\binom{14}{3}\binom{11}{3}\binom{9}{3}` ways of conforming all the groups.