Question

Evaluate ${}^4{C}_3{+}^4{C}_4$

Answer 1

We know that

${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$

Given ${}^4{C}_3{+}^4{C}_4$

Where $n=4,r=3,4$

$$\begin{gathered} {\therefore }^n{C}_r=\frac{n!}{r!\left(n-r\right)!} \\ {=}^4{C}_3{+}^4{C}_3 \\ =\frac{4!}{3!\left(4-3\right)!}+\frac{4!}{4!\left(4-4\right)!} \\ =\frac{4!}{3!\left(1\right)!}+\frac{4!}{4!\left(0\right)!} \\ but0!=1and1!=1 \\ =\frac{4!}{3!\times 1}+\frac{4!}{4!\times 1} \\ =\frac{4\times 3!}{3!\times 1}+\frac{4!}{4!} \\ =\frac{4}{1}+\frac{1}{1} \\ =4+1 \\ =5 \end{gathered}$$

Hence, ${}^4{C}_3{+}^4{C}_4$ is $5$