Question

evaluate each expression. sample problem ( _8p_3 ) ( \frac{8!}{(8-3)!} = \frac{8!}{5!} = 336 ) ( _5c_3 ) 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 \( _5C_3 \), we have \( n = 5 \) and \( r=3 \).

Step2: Substitute values into the formula

Substitute \( n = 5 \) and \( r = 3 \) into the formula: \( _5C_3=\frac{5!}{3!(5 - 3)!}=\frac{5!}{3!2!} \).

Step3: Expand the factorials

We know that \( n!=n\times(n - 1)\times\cdots\times1 \), so \( 5! = 5\times4\times3\times2\times1 \), \( 3! = 3\times2\times1 \), and \( 2! = 2\times1 \). Then the expression becomes \( \frac{5\times4\times3\times2\times1}{(3\times2\times1)\times(2\times1)} \).

Step4: Simplify the expression

Cancel out the common terms in the numerator and the denominator. The \( 3\times2\times1 \) in the numerator and denominator cancels out, leaving \( \frac{5\times4}{2\times1} \).

Step5: Calculate the result

Calculate \( \frac{5\times4}{2\times1}=\frac{20}{2}=10 \).

Answer:

10