# Factorization of combinatorics equation

I was watching a video on combinatorics and the lecturer presented the original answer and the factored(simplified) answer.

$${({{13}\choose{5}}-10)(4^5-4)}={{13}\choose{5}}4^5-{{13}\choose{5}}4-10(4^5+4)$$
I realised that my expanded equation is missing $$+ 10 * 4$$, where did I got my steps wrong?
• You have not expanded the $-10(4^5 - 4)$ term. Try that. – N. F. Taussig Apr 7 at 15:48
• if you also expand the factor $10(4^5 - 4)$ (you made a small mistake in your first equation), you'll find the original equation (if that is your question). The notice answer is obtained from reversing the step you take (after expanding said factor). – Student Apr 7 at 15:48
• I don't see how expanding $-10*4^5 -10*4$ gives me 10*4 = 40, which is part of the original equation (answer) – ilovetolearn Apr 7 at 15:55
• That $-10*4$ should be $+10*4$. You distribute $\binom{13}{5}$ over both terms in $(4^5 - 4)$. You don't do this for the $-10$, hence in the second right hand side of your equation, there hsould be $-10(4^5 \color{#c00}{-} 4)$ which expands to $-10*4^5 + 10*4$ as in the original equation. – Student Apr 7 at 17:03
• Starting from the original answer, factor out the $4^5$ from the first two terms and factor out a $-4$ from the last two terms. Now each term has a common factor of $\left({{13}\choose{5}} - 10\right)$ which you can factor out. – NickD Apr 7 at 17:07