Question 18.7: Suppose you have to form a team of five students from a grou...

a. Equation (18.35a) gives the number of ordered groups of R = 5 things chosen from a group of N = 10 as

P_R^N = \frac{N!}{(N − R)!}                (18.35a)

(P_5^{10})_{\substack{\text{using each}\\\text{student only once}\\}} = \frac{10!}{(10 − 5)!} = \frac{10!}{5!} = \frac{10 × 9 × 8 × 7 × 6 × 5!}{5!}

= 10 × 9 × 8 × 7 × 6 = 30,240  \text{groups}

b. Equation (18.35c) gives the result when the students are allowed to be in more than one group:

P_N^N = N^R            (18.35c)

(P_3^{10})_{\substack{\text{using each student}\\\text{more than once}\\}} = 10^5 = 100,000  \text{groups}

c. If the students are not assigned a position within the group, but they are allowed to belong to only one group, then Eq. (18.36a) gives the possible number of groups as

C_R^N = \frac{P_R^N}{R!} = \frac{N!}{(N −R)! R!}            (18.36a)

(C_5^{10})_{\substack{\text{using each}\\\text{student only once}\\}} = \frac{10!}{(10 − 5)! 5!} =  \frac{10 × 9 × 8 × 7 × 6 × 5!}{5! × 5! } = 252  \text{groups}

d. If the students are not assigned a position within the group, but they are allowed to belong to more than one group, then Eq. (18.36b) gives the possible number of groups as

C_R^N = \frac{P_R^N}{R!} = \frac{(N+R−1)!}{(N −1)! R!}              (18.36b)

(C_5^{10})_{\substack{\text{using each student}\\\text{more than once}\\}} = \frac{(10+5−1)!}{(10 − 1)! 5!} = \frac{14 × 13 × 12 × 11 × 10 × 9!}{9! × 5! } = 2002  \text{groups}

e. If there are four men and six women in the group, then the number of different unordered ten-person groups that can be formed is given by Eq. (18.37) as

P_{R_1,R_2 ,…,R_k}^N = \frac{N!}{R_1!R_2!R_3!…R_k!}                (18.37)

P_{4,6}^{10} = \frac{10!}{4! × 6!} = \frac{10 × 9 × 8 × 7 × 6!}{4 × 3 × 2 × 1 × 6!} = 210  \text{groups}