Solution
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given combination expression
[tex]^nC_4[/tex]
STEP 2:Write the formula for combination
[tex]^nC_r=\frac{n!}{(n-r)!r!}[/tex]
STEP 3: Substitute the values
[tex]^nC_4=\frac{n!}{(n-4)!4!}[/tex]
Rewrite n!
[tex]n!=n(n-1)(n-2)(n-3)(n-4)![/tex]
We stop at (n-4)! so that it can be used to cancel out the (n-4)! which is the denominator of the combination expression. Therefore, we have:
[tex]\begin{gathered} \frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!4!} \\ \\ (n-4)!\text{ cancels out each other to have:} \\ \frac{n(n-1)(n-2)(n-3)}{4!} \\ 4!=4\times3\times2\times1 \\ We\text{ have:} \\ \frac{n(n-1)(n-2)(n-3)}{4\times3\times2\times1}=\frac{n(n-1)(n-2)(n-3)}{24} \end{gathered}[/tex]
Hence, the reason for having the expression in the image question.
