Question

a group consists of seven democrats and five republicans. four people are selected to attend a conference.
a. in how many ways can four people be selected from this group of twelve?
b. in how many ways can four republicans be selected from the five republicans?
c. find the probability that the selected group will consist of all republicans.

a. the number of ways to select four people from the group of twelve is 495.
b. the number of ways to select four republicans from the group of five republicans is \\(\square\\).

Explanation:

Step1: Recall Combination Formula

The number of ways to choose \( r \) items from \( n \) items is given by the combination formula \( C(n, r)=\frac{n!}{r!(n - r)!} \), where \( n!=n\times(n - 1)\times\cdots\times1 \) and \( 0!=1 \).

Step2: Apply Formula for Part b

Here, \( n = 5 \) (total Republicans) and \( r = 4 \) (Republicans to be selected). So we calculate \( C(5, 4)=\frac{5!}{4!(5 - 4)!} \).

First, expand the factorials: \( 5! = 5\times4\times3\times2\times1 \), \( 4! = 4\times3\times2\times1 \), and \( (5 - 4)!=1!=1 \).

Substitute into the formula: \( C(5, 4)=\frac{5\times4!}{4!\times1!} \). The \( 4! \) terms in the numerator and denominator cancel out, leaving \( \frac{5}{1}=5 \).

Answer:

5