Question

1. evaluate $^4c_2$.

$^4c_2 = $ enter your next step here

Explanation:

Step1: Recall the combination formula

The formula for combinations is \( _nC_r=\frac{n!}{r!(n - r)!} \). For \( _4C_2 \), we have \( n = 4 \) and \( r=2 \).

Step2: Substitute values into the formula

Substitute \( n = 4 \) and \( r = 2 \) into the formula: \( _4C_2=\frac{4!}{2!(4 - 2)!}=\frac{4!}{2!2!} \).

Step3: Expand the factorials

We know that \( n!=n\times(n - 1)\times\cdots\times1 \). So, \( 4!=4\times3\times2\times1 = 24 \), \( 2!=2\times1=2 \). Then \( \frac{4!}{2!2!}=\frac{24}{2\times2} \).

Step4: Perform the division

Calculate \( \frac{24}{4}=6 \).

Answer:

\( _4C_2 = 6 \)