Question

in how many ways can a committee of three democrats and four republicans be formed from a group of six democrats and eleven republicans? a committee of three democrats and four republicans can be formed from a group of six democrats and eleven republicans in \\(\square\\) different ways.

Explanation:

Step1: Calculate Democrat combinations

The number of ways to choose 3 Democrats from 6 is given by the combination formula $C(n,k)=\frac{n!}{k!(n-k)!}$. For Democrats: $C(6,3)=\frac{6!}{3!(6-3)!}=\frac{6×5×4}{3×2×1}=20$.

Step2: Calculate Republican combinations

For Republicans, choose 4 from 11: $C(11,4)=\frac{11!}{4!(11-4)!}=\frac{11×10×9×8}{4×3×2×1}=330$.

Step3: Multiply the two results

Total ways = Democrats combinations × Republicans combinations = $20×330=6600$.

Answer:

6600