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Question
- committee representation there are 6 republican, 5 democrat, and 4 independent candidates. how many different ways can a committee of 3 republicans, 2 democrats, and 1 independent be selected?
Step1: Calculate the number of ways to choose Republicans
Use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 6\) (number of Republican candidates) and \(r=3\) (number of Republicans to be chosen).
$$C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6\times5\times4\times3!}{3!\times3!}=\frac{6\times5\times4}{3\times2\times1}=20$$
Step2: Calculate the number of ways to choose Democrats
Use the combination formula with \(n = 5\) (number of Democrat candidates) and \(r = 2\) (number of Democrats to be chosen).
$$C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4\times3!}{2!\times3!}=\frac{5\times4}{2\times1}=10$$
Step3: Calculate the number of ways to choose Independents
Use the combination formula with \(n = 4\) (number of Independent candidates) and \(r = 1\) (number of Independents to be chosen).
$$C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4\times3!}{1!\times3!}=4$$
Step4: Use the multiplication principle
Multiply the number of ways of choosing Republicans, Democrats, and Independents together.
$$N = C(6,3)\times C(5,2)\times C(4,1)=20\times10\times4$$
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