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The principal would like to assemble a committee of 7 students from the 12 -member student council. How many different committees can be chosen?
Answer:

The principal would like to assemble a committee of $7$ students from the $12$ -member student council. How many different committees can be chosen? $\newline$ Answer:

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Full solution

Q. The principal would like to assemble a committee of

7

students from the

12

-member student council. How many different committees can be chosen?

\newline

Answer:

Identify Problem Type: Identify the type of problem. $\newline$ We need to find the number of ways to choose $7$ students out of $12$ without regard to the order in which they are chosen. This is a combination problem, not a permutation, because the order does not matter.
Use Combination Formula: Use the combination formula. $\newline$ The number of combinations of $n$ items taken $k$ at a time is given by the formula: $\newline$ $C(n, k) = \frac{n!}{k!(n - k)!}$ $\newline$ where $n!$ denotes the factorial of $n$ , which is the product of all positive integers up to $n$ .
Apply Formula: Apply the formula to our problem. $\newline$ Here, $n = 12$ (the total number of students) and $k = 7$ (the number of students to choose for the committee). $\newline$ $C(12, 7) = \frac{12!}{7!(12 - 7)!}$
Simplify Expression: Simplify the expression. $\newline$ $C(12, 7) = \frac{12!}{7! \times 5!}$ $\newline$ $C(12, 7) = \frac{(12 \times 11 \times 10 \times 9 \times 8)}{(5 \times 4 \times 3 \times 2 \times 1)}$
Perform Calculation: Perform the calculation. $\newline$ $C(12, 7) = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}$ $\newline$ $C(12, 7) = \frac{12}{1} \times \frac{11}{1} \times \frac{10}{2} \times \frac{9}{3} \times \frac{8}{4}$ $\newline$ $C(12, 7) = 12 \times 11 \times 5 \times 3 \times 2$ $\newline$ $C(12, 7) = 12 \times 11 \times 5 \times 3 \times 2$ $\newline$ $C(12, 7) = 12 \times 11 \times 30$ $\newline$ $C(12, 7) = 3960$

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