1. combinations

(1)There are 10 people who are going to spend the afternoon in 2 groups , one group goes to the theater and the other plays tennis . In how many ways can the group for tennis be selected if there must be at least 4 people in each group ?

is it (4x6)+(5x5)+(6x4)=73 ??

(2) A committe consisting of 2 men and 2 women is to be chosen from 5 men and 6 women . One of the women is the wife of one of the man . In how many ways can the committe be chosen if it must contain either the man and his wife or neither .

Not sure bout this >

2. 2)

So there are two cases,

1. Both the man and wife are present.

In that case, two members are already fixed, so you need to select the remaining two, one man from 4 and one woman from 5 , so the number of combinations =20

2. Neither are on the committee.

Here you need to select two men out of 4 and two women out of 5, so no. of possibilities = 6(10) = 60

So total number of ways = 60+20 = 80

3. 1)

There are three situations,

1. There are 4 people in the tennis group

2. There are 5 people in the tennis group.

3. There are 6 people in the tennis group.

Total number of combinations = $\binom{10}{4}+\binom{10}{5}+\binom{10}{6}$

= 420 + 252 = 672 ??

(I'm not sure of the method.)