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This is from Serge Lang's book Basic Mathematics. I've stuck at this for over six hours now. Read multiple posts, forums, and even checked the answer key.

The main thing I'm confused about is how can we get this common denominator: $n!(m - n + 1)!$

Everything else is easy.

Théophile
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Roarke
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    What is the relation between $(m-n)!$ and $(m-n+1)!$ ? Another thing to try: Do it with ${5\choose 3}+{5\choose 2}$ and see what happens. – GEdgar Mar 18 '21 at 22:04
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    This is called Pascal's identity. That should allow you to research it. – Favst Mar 18 '21 at 22:06
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    The calculations you've tried aside, would you be interested in a combinatorial proof? There will be a question for that already as well. – J.G. Mar 18 '21 at 22:14
  • Yes, I'd be interested in knowing that as well. – Roarke Mar 18 '21 at 22:36
  • It answers the question, yes. But I'm not really trying to find an answer, I'm trying to understand the concept. And it's starting to make sense now. – Roarke Mar 18 '21 at 22:37

2 Answers2

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You have $m + 1$ biscuits and you want to pick $n$ of them.

Put butter on one of the biscuits. Now to pick $n$, you can either pick the biscuit with the butter, and then you still have to pick $n-1$ more (a total of $m \choose n-1$ ways to do this), or you can avoid the biscuit with the butter, in which case you have to pick all $n$ from the remaining $m$ (a total of $m \choose n$ ways to do this).

hunter
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Quite simple: \begin{align} \binom mn+\binom m{n-1}&=\frac{m!}{n!(m-n)!}+\frac{m!}{(n-1)!(m-n+1)!}\\ & =\frac{m!}{\color{red}n\,(n-1)!(m-n)!}+\frac{m!}{(n-1)!(m-n)!\,(\color{blue}{m-n+1})} \\ &=\frac{[(\color{blue}{m-n+1})+\color{red}n]m!}{n!\,(m-n+1)!}=\frac{(m+1)!}{n!\,(m-n+1)!}. \end{align}

Bernard
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  • Blue might be easier to read than cyan. – J.G. Mar 18 '21 at 22:24
  • You're probably right. I'll fix it. B.t.w. thanks for the edit! – Bernard Mar 18 '21 at 22:25
  • @Bernard Can you explain this part again?

    $\frac{[(\color{blue}{m-n+1})+\color{red}n]m!}{n!,(m-n+1)!}=\frac{(m+1)!}{n!,(m-n+1)!}$

    I don't understand what happened here. The colored part, how did that become $(m+1)!$ ? Did you just cancel out the positive and negative $n$? But then $m!$ would still be there, right? I'm a little overwhelmed now.

     – Roarke Mar 18 '21 at 23:00
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    Yes, in the reduction to the same denominator, the $n$s cancel, and there remains $(m+1),m!$, which is $(m+1)!$ by the recursive definition of the factorial. – Bernard Mar 18 '21 at 23:04
  • I'd like to know one more thing. In the end, you've written $n!(m-n+1)!$ in that, where does the $(m-n+1)$ part goes? Because we have to show that the denominator is equal to $n!$ only, right? I am assuming that you expanded $(m+1)!$ so that it becomes $(m-n)!\cdot (m-n+1)$ and then cancel $(m-n+1)!$ out. Am I getting it correctly? Thanks! – Roarke Mar 19 '21 at 00:23
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    I didn't expand $(m+1)!$, and the final denominator is $n!,(m-n+1)!$ (the l.c.m. of the initial denominators). I coloured the factor od each denominator with which we have to multiply the other fraction (both numerator and denominator) to obtain the same final denominator for each fraction. Probably you should redo it by yourself to have a complete feeling of how it works. – Bernard Mar 19 '21 at 00:34