How do you prove the binomial coefficient identity and compute $\binom{60}{55}$?

Key takeaway: This is a classic algebra-with-factorials problem. The proof is mostly about rewriting the factorials carefully, and the final calculation uses the identity in a very efficient way.

(a) Show that

$3\left(\frac{5}{2}\right)=5\left(\frac{4}{2}\right).$

Using the numbers exactly as written:

$3\left(5\times\frac{4}{2}\right)=5\left(\frac{4\times3}{2}\right).$

Now simplify both sides:

$3(5\cdot2)=5(2\cdot3).$

So both sides equal

$30.$

(b) Show that

$(n-k)\binom{n}{k}=n\binom{n-1}{k}, \qquad n>k.$

Start from the left-hand side:

$(n-k)\binom{n}{k}=(n-k)\frac{n!}{k!(n-k)!}.$

Since $n!=(n)(n-1)!$, we get

$(n-k)\binom{n}{k}=(n-k)\frac{n(n-1)!}{k!(n-k)!}.$

Cancel one factor of $(n-k)$:

$(n-k)\binom{n}{k}=\frac{n(n-1)!}{k!(n-k-1)!}.$

But

$\binom{n-1}{k}=\frac{(n-1)!}{k!(n-1-k)!},$

so the expression becomes

$n\binom{n-1}{k}.$

Hence,

$(n-k)\binom{n}{k}=n\binom{n-1}{k}.$

(c) Evaluate $\binom{60}{55}$ given $\binom{59}{55}=455126$.

Use the identity with $n=60$ and $k=55$:

$(60-55)\binom{60}{55}=60\binom{59}{55}.$

So

$5\binom{60}{55}=60\times455126.$

Divide by 5:

$\binom{60}{55}=12\times455126=5461512.$

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Pitfalls the pros know 👇 In part (b), the dangerous step is canceling factorial terms too quickly and losing track of the remaining factor. Write out the factorials before simplifying.

In part (c), another common mistake is to forget that $60-55=5$, so you must divide by 5 at the end. Also, remember symmetry: $\binom{60}{55}=\binom{60}{5}$, which can be a helpful check.

What if the problem changes? If the given value were $\binom{59}{54}$ instead, the same identity could be used with $k=54$. More generally,

$(n-k)\binom{n}{k}=n\binom{n-1}{k}$

lets you move between neighboring rows of Pascal’s triangle without expanding every factorial from scratch.

Tags: binomial coefficient, factorial identity, Pascal triangle