Question

evaluate each expression.
sample problem
$_8p_3$
$\frac{8!}{(8-3)!} = \frac{8!}{5!} = 336$
$_7c_2$
enter the answer in the space provided. use numbers instead of words.

Explanation:

Step1: Recall the combination formula

The formula for combinations is \( _nC_r=\frac{n!}{r!(n - r)!} \). For \( _7C_2 \), we have \( n = 7 \) and \( r=2 \).

Step2: Substitute into the formula

Substitute \( n = 7 \) and \( r = 2 \) into the formula: \( _7C_2=\frac{7!}{2!(7 - 2)!}=\frac{7!}{2!×5!} \).

Step3: Expand the factorials

We know that \( n!=n\times(n - 1)\times\cdots\times1 \), so \( 7! = 7\times6\times5! \). Then the expression becomes \( \frac{7\times6\times5!}{2!×5!} \).

Step4: Simplify the expression

The \( 5! \) terms in the numerator and denominator cancel out. \( 2! = 2\times1=2 \), so we have \( \frac{7\times6}{2\times1}=\frac{42}{2}=21 \).

Answer:

21