Question

A country has two political parties, the Demonstrators and the Repudiators. Suppose that the national senate consists of 100 members, 44 of which are Demonstrators and 56 of which are Repudiators.
(a) How many ways are there to select a committee of 10 senate members with the same number of Demonstrators and Repudiators?
(b)Suppose that each party must select a speaker and a vice speaker. How many ways are there for the two speakers and two vice speakers to be selected?

Answer 1

Answer:

(a) 4.148 x 10^(12) ways

(b) 5,827,360 ways

Step-by-step explanation:

Number of Demonstrators (D)  = 44

Number of Repudiators (R) = 56

(a)

5 senate members must be Repudiators and 5 must be demonstrators, assuming that the order at which they are selected is irrelevant:

$N= C^{D}_{5} * C^{R}_{5}\\N=\frac{56!}{5!(56-5)!} *\frac{44!}{5!(44-5)!} \\N=3,819,816*1,086,008\\N=4.148 *10^{12}$

(b)

Since there are two different positions, (speaker and vice speaker), order is important in this situation, and the total number of ways to select two senators from each party is:

$N= P^{D}_{2} * P^{R}_{2}\\N=\frac{56!}{(56-2)!} *\frac{44!}{(44-2)!} \\N=3,080*1,892\\N=5,827,360$