1
$\begingroup$

This is an equation from my textbook that I am trying to understand:

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

What I got so far, is that $\frac{k\cdot n!}{k!} = \frac{n!}{(k-1)!}$ however, why does the same principle apply for (n-k)! in the denominator? Isn't there only one k in the numerator that I can cancel out in the denominator?

In other words, shouldn't it be $\frac{k\cdot n!}{k!(n-k)!} = \frac{n(n-1)!}{(k-1)!((n-k)!}$ ?

$\endgroup$
4
$\begingroup$

Hint: Simple! use the following simple equations: $$(n-1) - (k-1) = n - k$$ And: $$n! = n \times (n-1)!$$

$\endgroup$
3
$\begingroup$

All you have to do is to rewrite $\dfrac{k}{k!}$ as $\dfrac{1}{(k-1)!}$ and note that $n! = n(n-1)!$ and $(n-1)-(k-1) = n-k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.