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(m, n) + (m, n-1) = (m +1, n)

I have read as many explanations of this problem that I could find and I still have no clue how to do it. I understand that m! can be written as m(m-1)!, and likewise for the rest of the problem but I don't know how to get the common denominator that is listed in the answer for this problem in the back of the book: n!(m−n+n)! . I am tearing my hair out over this and would appreciate any guidance. Thanks.

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    This is Pascal's rule. https://en.wikipedia.org/wiki/Pascal%27s_rule – CyclotomicField Nov 21 '23 at 00:17
  • Welcome to the math SE. posts here are much better received if they provide details of the efforts one has made to solve their problems themselves. You write that you have read explanations. What explanations have you read, and what is unclear about them? – Paul Tanenbaum Nov 21 '23 at 00:19
  • One explanation I have read and not understood was this answer which was given to another math SE user who was having the same problem: https://math.stackexchange.com/a/4067523/1255275 I understand steps 1 and 2 but I am lost where he brings the highlighted sections into the numerator –  Nov 21 '23 at 00:26
  • @CalL. Adding fractions with different denominators, $$\frac1A + \frac1B = \frac{B}{AB} + \frac{A}{AB} = \frac{A+B}{AB}$$ – peterwhy Nov 21 '23 at 00:42
  • after expanding the denominators I realized that I could find a common denominator by multiplying the first coefficient by (m-n+1), and the second by n. I am left with the denominator n!(m-n+1) which is what is listed in the back of the book, but I don't understand how the numerator m!(m-n+1) + m!n turns into m!(m+1). I understand why the -n is cancelled out but where did the other m! go? –  Nov 21 '23 at 00:59
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    @CalL. For the numerator $m! \color{blue}{(m-n+1)} + m!\color{red}n$, notice the common factor $m!$ to get $m![\color{blue}{(m-n+1)}+\color{red}n]$. Then continue with "the -n is cancelled out" as in your comment. – peterwhy Nov 21 '23 at 01:13
  • @peterwhy. That makes perfect sense, I am kicking myself over making such a simple mistake. Thanks for your patience. –  Nov 21 '23 at 01:21
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    @CalL. Welcome, but please don't kick yourself and don't tear your hair out. – peterwhy Nov 21 '23 at 01:24

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