Combinatorics Question. Id like a thorough explanation. Especially the part about multiplying or dividing each side and when its OK to do so.

Question

The question is : ${n \choose 3} +{n \choose 2}+{n \choose 1}$. and it says to write the following expression without using the factorial symbol. Thanks in advance.

The answer I got was $n^4 - \frac {3n}{6} + n^2 - \frac n2 + \frac n1$

Answer 1

The notation $n C k$, read "$n$ choose $k$" and alternatively denoted $\binom{n}{k}$, means: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ So then $$\begin{align*} \binom{n}{3} + \binom{n}{2} + \binom{n}{1} &= \frac{n!}{3!(n-3)!} + \frac{n!}{2!(n-2)!} + \frac{n!}{1!(n-1)!} \\ &= \frac{1}{6}n(n-1)(n-2) + \frac{1}{2}n(n-1) + n \\ &= \frac{1}{6}n^3 - \frac{3}{6}n^2 + \frac{2}{6}n + \frac{1}{2}n^2 - \frac{1}{2}n + n \\ &= \frac{1}{6}n^3 + \frac{5}{6}n \end{align*}$$ I believe the answer you gave was incorrect.