Question

substitute for n and r and simplify the expression.

\\(_{n}c_{r} = \frac{n!}{(n-r)!r!}\\)

\\(_{47}c_{8} = \frac{47!}{(47-8)!8!}\\) substitute.

\\(= \frac{47!}{(39)!8!}\\) subtract.

simplify \\(\frac{47!}{39!8!}\\) to find \\(_{47}c_{8}\\). recall that \\(n! = n(n-1)(n-2)\cdots(3)(2)(1)\\).

\\(_{47}c_{8} = \frac{47!}{39!8!}\\)
\\(= \square\\) (simplify your answer. type an integer or a fraction.)

Explanation:

Set up the combination formula

$$ _{47}C_{8} = \frac{47!}{39! \cdot 8!} $$

Expand and simplify the factorials

$$
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$$

Calculate the final integer value

$$ _{47}C_{8} = 314,457,495 $$

Answer:

Simplify \(\frac{47!}{39!8!}\) to find \(_{47}C_{8}\). Recall that \(n! = n(n-1)(n-2)\cdots(3)(2)(1)\).

\(_{47}C_{8} = \frac{47!}{39!8!}\)
\(=\) <blank>314457495</blank> (Simplify your answer. Type an integer or a fraction.)