I\'m not sure how to do f. As for as a to e I got: a) 396 ways b) 142,506 ways c

Question

I'm not sure how to do f. As for as a to e I got:

a) 396 ways

b) 142,506 ways

c) 3,769,920 ways

d) 34,902 ways

e) 215.433 x 10^18 ways

If you could do them all, so I can verify my values and learn how to do f. Please show all work.

A class contains 22 girls and 18 boys. For all parts of this question, each boy and girl are distinguishable from one another. Answer the following questions: In how many ways can a committee of one boy and one girl be chosen? In how many ways can a committee of five students be chosen? In how many ways can a committee of two girls and three boys be chosen? In how many ways can a committee of five students be chosen such that all the students on the committee are the same sex? In how many ways can the girls and boys form a line where no two boys are standing next to one another? How many committees of five students contain at least two girls?

Explanation / Answer

We can solve by two methods :

(1) Inclusion exclusion Principle

Number of committees with 0 girl = 18c5 = 8568 ways

Number of committees with 1 girl = 18c4 * 22c1 = 67320ways

Total ways = 40c5 = 658008 ways

Thus, commitees with atleast 2 girls = 658008 - 8568 - 67320 = 582120

(2) General Ovservation Method.

Since we want atleast 2 girls so we start selecting 5 members as :

2 girls and 3 boys = 22c2*18c3

3 girls 2 boys = 22c3*18c2

4 girls 1 boy = 22c4*18c1

5 girls 0 boys = 22c5*18c0

On summing up the above results we get : 582120

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