Question

a) In how many ways can a committee of 5 be chosen from 9 people?

(b) How many committees of 5 or more can be chosen from 9 people?

(c) In how many ways can a committee of 5 teachers and 4 students be chosen

from 9 teachers and 15 students?

(d) In how many ways can the committee in (C) be formed if teacher A refuses

to serve if student B is on the committee?

Answer 1

a) If the order matters (we distinguish people of a committee)

$$\dfrac{9!}{(9-5)!}=9(8)(7)(6)(5)=15120$$

If the order does not matter

$$\dbinom{9}{5}=\dfrac{9!}{5!(9-5)!}=\dfrac{9(8)(7)(6)(5)}{120}=126$$

(b) If the order matters

$$\dfrac{9!}{(9-5)!}+\dfrac{9!}{(9-6)!}+\dfrac{9!}{(9-7)!}$$

$$+\dfrac{9!}{(9-8)!}+\dfrac{9!}{(9-9)!}$$

$$=15120+60480+181440+362880+362880$$

$$=982800$$

If the order does not matter

$$\dbinom{9}{5}+\dbinom{9}{6}+\dbinom{9}{7}+\dbinom{9}{8}+\dbinom{9}{9}$$ $$=126+84+36+9+1=256$$

(c)

$$\dbinom{9}{5}\dbinom{15}{4}=\dfrac{9!}{5!(9-5)!}\cdot \dfrac{15!}{4!(15-4)!}$$

$$=126\cdot1365=171990$$

(d)

In how many ways can a committee of 5 teachers and 4 students be chosen

from 9 teachers and 15 students if  teacher A and student B are on the committee

teacher A: $\dbinom{8}{4}$

student B: $\dbinom{14}{3}$

 teacher A and student B $\dbinom{8}{4}\dbinom{14}{3}$

Then the number of ways can the committee in (C) be formed if teacher A refuses to serve if student B is on the committee is

$$\dbinom{9}{5}\dbinom{15}{4}-\dbinom{8}{4}\dbinom{14}{3}$$

$$=171990-70(364)=146510$$