How many different ways can a committee of 3 republicans, 2 democrats and 1 independent be selected **Problem Statement:***Combinatorial Selection Problem*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?**Explanation:**This problem deals with selecting a specific number of candidates from different groups. It can be solved using combinations, denoted as \( C(n, k) \) or \( \binom{n}{k} \), which represents the number of ways to choose \( k \) items from \( n \) items without regard to order.1. **Number of Ways to Choose 3 Republicans from 6:** \[ C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]2. **Number of Ways to Choose 2 Democrats from 5:**\[ C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]3. **Number of Ways to Choose 1 Independent from 4:**\[ C(4, 1) = \frac{4!}{1!(4-1)!} = \frac{4}{1} = 4 \]**Total Number of Ways to Form the Committee:**To find the total number of ways to form the committee, multiply the number of ways to select each group:\[ 20 \times 10 \times 4 = 800 \]Therefore, there are 800 different ways to select a committee of 3 Republicans, 2 Democrats, and 1 Independent.

Name: How many different ways can a committee of 3 republicans, 2 democrats
Uploaded: 2024-03-05T11:53:49.000Z
Channel: Bartleby
Description: How many different ways can a committee of 3 republicans, 2 democrats and 1 independent be selected **Problem Statement:***Combinatorial Selection Problem*There are 6 Republican, 5 Democrat, and 4 Independent candidates. How many different ways can a