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

a. select one boy from 18 boys and one girl from 22 girls so=> 18C1 * 22C1= 18 * 22 =396

b. commitee of 5 student can be chossed from 40(18+22) available is =22C5 = 22!/5! * 17! = 22*21*20*19*18/120

=>22*21*19*18/6=22*21*19*3 = 462*57=26334

c. select 2 girls from 22 girlas and 3 boys from 18 boys = 22C2 * 18C3 = 22!/2!*20! * 18!/3!*15!

11*21 * 3*17*16=206976

d. either select 5 boys from 18 boys or select 5 girls from 22 girls = 18C5+22C5 =225629712

e. first arrange girls in 22! ways now we have 23 gaps available now select 18 gaps from 23 available gaps and arrange boys in 18! ways

22! * 23C18 * 18!(calculate it)

f. here 2G3B or 3G2B OR 4G1B OR 5G

SO 22C2*18C3 + 22C3*18C2 + 22C4*18C1 + 22C5 (Calculate it)

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