Solution
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given combination expression
![^nC_4](https://img.qammunity.org/2023/formulas/mathematics/high-school/u7u2k0w65nln1t7gc8lnx0sxpnlwabolau.png)
STEP 2:Write the formula for combination
![^nC_r=(n!)/((n-r)!r!)](https://img.qammunity.org/2023/formulas/mathematics/college/1q6cebegtgha4ctlij3deo1h0foedhbfzo.png)
STEP 3: Substitute the values
![^nC_4=(n!)/((n-4)!4!)](https://img.qammunity.org/2023/formulas/mathematics/high-school/4z7dhi6fr4k41psvzmh4a8mosrq75pvzb3.png)
Rewrite n!
![n!=n(n-1)(n-2)(n-3)(n-4)!](https://img.qammunity.org/2023/formulas/mathematics/high-school/q0nzdzlrs6fgyajjvcvpgt3um3t3sevwt7.png)
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:
![\begin{gathered} (n(n-1)(n-2)(n-3)(n-4)!)/((n-4)!4!) \\ \\ (n-4)!\text{ cancels out each other to have:} \\ (n(n-1)(n-2)(n-3))/(4!) \\ 4!=4*3*2*1 \\ We\text{ have:} \\ (n(n-1)(n-2)(n-3))/(4*3*2*1)=(n(n-1)(n-2)(n-3))/(24) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/lini55tca620w5t2deq4gfttmfdzbkliek.png)
Hence, the reason for having the expression in the image question.