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How many ways can a delegation of 5 republicans, 5 Democrats, and 1 independent be elected from a group of 6 republicans, 6 democrats, and 2 independents?

User Vasspilka
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Final answer:

The total number of ways a delegation of 5 Republicans, 5 Democrats, and 1 Independent can be elected from a group of 6 Republicans, 6 Democrats, and 2 Independents is 72.

Step-by-step explanation:

To find the number of ways a delegation can be elected, we need to use the concept of combinations. We have 6 Republicans, 6 Democrats, and 2 Independents, and we need to select 5 Republicans, 5 Democrats, and 1 Independent.

The number of ways to select 5 Republicans from 6 is denoted as C(6, 5) = 6.

The number of ways to select 5 Democrats from 6 is denoted as C(6, 5) = 6.

The number of ways to select 1 Independent from 2 is denoted as C(2, 1) = 2.

By using the Multiplication Principle, we can multiply the number of ways for each group together:

Total number of ways = C(6, 5) * C(6, 5) * C(2, 1) = 6 * 6 * 2 = 72.

Therefore, there are 72 ways a delegation of 5 Republicans, 5 Democrats, and 1 Independent can be elected from the given group.

User Leopardxpreload
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