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Evaluate $\newline$ $_{9}C_{6}$ and $\newline$ $_{11}P_{4}$ . $\newline$ \begin{array}{l} _{9}C_{6}=\Box,\ _{11}P_{4}=\Box \end{array}

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Add and subtract three or more integers

Full solution

Q. Evaluate

\newline

_{9}C_{6}

and

\newline

_{11}P_{4}

.

\newline

\begin{array}{l} _{9}C_{6}=\Box,\ _{11}P_{4}=\Box \end{array}

Evaluate Combination: Evaluate $_(9)C_(6)$ , which represents the number of combinations of $9$ items taken $6$ at a time. $\newline$ The formula for combinations is: $\newline$ $_(n)C_(r) = \frac{n!}{r!(n - r)!}$ $\newline$ where $n!$ denotes the factorial of $n$ . $\newline$ Let's calculate $_(9)C_(6)$ using the formula. $\newline$ $_(9)C_(6) = \frac{9!}{6!(9 - 6)!}$
Simplify Combination: Simplify the expression for $_{9}C_{6}$ . $_{9}C_{6} = \frac{9!}{6!3!}$ $9! = 9 \times 8 \times 7 \times 6!$ $3! = 3 \times 2 \times 1$ Now, we can cancel out the common $6!$ term in the numerator and denominator. $_{9}C_{6} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1}$
Calculate Combination: Perform the calculation for $_(9)C_(6)$ .
$_(9)C_(6) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1}$
$_(9)C_(6) = \frac{504}{6}$
$_(9)C_(6) = 84$
Evaluate Permutation: Evaluate $_{11}P_{4}$ , which represents the number of permutations of $11$ items taken $4$ at a time. $\newline$ The formula for permutations is: $\newline$ $_{n}P_{r} = \frac{n!}{(n - r)!}$ $\newline$ Let's calculate $_{11}P_{4}$ using the formula. $\newline$ $_{11}P_{4} = \frac{11!}{(11 - 4)!}$
Simplify Permutation: Simplify the expression for $_{11}P_{4}$ . $_{11}P_{4} = \frac{11!}{7!}$ $11! = 11 \times 10 \times 9 \times 8 \times 7!$ Now, we can cancel out the common $7!$ term in the numerator and denominator. $_{11}P_{4} = 11 \times 10 \times 9 \times 8$
Calculate Permutation: Perform the calculation for $(_{11}P_{4})$ . $\newline$ $(_{11}P_{4}) = 11 \times 10 \times 9 \times 8$ $\newline$ $(_{11}P_{4}) = 7920$

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