How do you simplify this equation:
$\dbinom{n}{k}-\dbinom{n-1}{k-1}-\dbinom{n-2}{k-1}$
I simplified it to this:
$\dfrac{n!}{k!(n-k)!}-\dfrac{(n-1)!}{(k-1)!(n-k)!}-\dfrac{(n-2)!}{(k-1)!(n-k-1)!}$
but I'm stuck on what to do next.
$\endgroup$3 Answers
$\begingroup$Using Pascal's Indentity twice:
$\begin{eqnarray} \dbinom{n}{k}-\dbinom{n-1}{k-1}-\dbinom{n-2}{k-1}&=&\dbinom{n}{k}-\left[\dbinom{n-1}{k-1}+\dbinom{n-2}{k-1}+\dbinom{n-2}{k}\right]+\dbinom{n-2}{k}\\ &=&\dbinom{n}{k}-\left[\dbinom{n-1}{k-1}+\dbinom{n-1}{k}\right]+\dbinom{n-2}{k}\\ &=&\dbinom{n}{k}-\dbinom{n}{k}+\dbinom{n-2}{k}\\ &=&\dbinom{n-2}{k}\\ \end{eqnarray}$
$\endgroup$ $\begingroup$Hint: Try breaking down the factorials to the least one of each type and then working with the fractions. For example, you have $n!$, $(n-1)!$, and $(n-2)!$. Break $n!$ down to $n(n-1)(n-2)!$ and $(n-1)!$ down to $(n-1)(n-2)!$. Do the same for the other types. This will help you get common denominators and help with simplifying.
$\endgroup$ $\begingroup$From the Pascal's triangle, $$\dbinom{n}{k}=\dbinom{n-1}{k}+\dbinom{n-1}{k-1}$$ and$$\dbinom{n-1}{k} = \dbinom{n-2}{k-1}+\dbinom{n-2}{k}$$ $$\rightarrow \dbinom{n}{k}-\dbinom{n-1}{k-1}-\dbinom{n-2}{k-1}=\dbinom{n-1}{k}-\dbinom{n-2}{k-1} = \dbinom{n-2}{k}$$
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