Question

in how many ways can a committee of two democrats and three republicans be formed from a group of ten democrats and nine republicans? a committee of two democrats and three republicans can be formed from a group of ten democrats and nine republicans in different ways

Explanation:

Step1: Calculate the number of ways to choose Democrats

The number of ways to choose \(2\) Democrats out of \(10\) is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 10\) and \(r=2\).

$$C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9\times8!}{2\times1\times8!}=45$$

Step2: Calculate the number of ways to choose Republicans

The number of ways to choose \(3\) Republicans out of \(9\) is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 9\) and \(r = 3\).

$$C(9,3)=\frac{9!}{3!(9 - 3)!}=\frac{9\times8\times7\times6!}{3\times2\times1\times6!}=84$$

Step3: Use the multiplication principle

By the multiplication principle (if one event can occur in \(m\) ways and another independent event can occur in \(n\) ways, then the two events together can occur in \(m\times n\) ways), the total number of ways to form the committee is \(C(10,2)\times C(9,3)\).

$$45\times84 = 3780$$

Answer:

\(3780\)