I'm finding it hard to understand some transitions in the solutions of my book. Is math.stackexchange a place to ask for help about this ?

This is the solution: The part I understood was only the first equaility, from there I'm lost..

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  • $\begingroup$ Common denominators. $\endgroup$
    – Randall
    Sep 2, 2017 at 11:50
  • $\begingroup$ Do you know what the common denominator is between $\frac1{(i-1)!}$ and $\frac1{i!}$ if you're noy sure try few $i$'s like $i=2,3,4,5,6$ and/or remember the definition of factorials. $\endgroup$
    – kingW3
    Sep 2, 2017 at 11:56

1 Answer 1


Start with the left hand side: $$ \binom {n-1}{i-1}+ \binom {n-1}i $$ expand it by definition : $$ \color{red}{\binom{n-1}{i-1}} + \color{green}{\binom{n-1}i} = \color{red}{\frac{(n-1)!}{(i-1)!(n-i)!}} + \color{green}{\frac{(n-1)!}{i!(n-i-1)!}} $$ where the colored part on the RHS corresponds to the coloured part on the LHS, by definition.

On the term on the RHS coloured red, multiply on the top and bottom by $i$. $$ \frac{(n-1)!}{(i-1)!(n-i)!} = \frac{i(n-1)!}{\color{blue}{i(i-1)!}(n-i)!} = \frac{i(n-1)!}{\color{blue}{i!}(n-i)!} $$ where I've highlighted the fact that $i(i-1)! = i!$ by the definition of factorial.

Something similar is done with the green term : we multiply the top and bottom by $(n-i)$ on this occasion : $$ \frac{(n-1)!}{i!(n-i-1)!} = \frac{(n-i)(n-1)!}{\color{brown}{(n-i)(n-i-1)!}i!} = \frac{(n-i)(n-1)!}{\color{brown}{(n-i)!}i!} $$

where I've highlighted the fact that $(n-i)(n-i-1)! = (n-i)!$ by the definition of factorial.

So our original red and green terms look different now: $$ \color{red}{\frac{(n-1)!}{(i-1)!(n-i)!}} + \color{green}{\frac{(n-1)!}{i!(n-i-1)!}} = \color{red}{\frac{i(n-1)!}{i!(n-i)!}} + \color{green}{\frac{(n-i)(n-1)!}{(n-i)!i!}} $$

Now, what do we know about adding fractions? We know that they must be brought to a common denominator, after which we can just add the numerators. Now, look at the coloured fractions above. What do you see? Both of them have the same denominator, which is $(n-i)!i!$, so you can directly add the numerators of these fractions. $$ {\frac{i(n-1)!}{i!(n-i)!}} + {\frac{(n-i)(n-1)!}{(n-i)!i!}} = {\frac{i{(n-1)!} + (n-i){(n-1)!}}{(n-i)!i!}} $$

Now, just focus on the numerator and how it simplifies: $$ i(n-1)! + \color{orange}{(n-i)(n-1)!} = i(n-1)! + \color{orange}{n(n-1)! - i(n-1)!} \\ \color{pink}{i(n-1)!} + n(n-1)! \color{pink}{- i(n-1)!} = n(n-1)! \\ n(n-1)! = n! $$

On the first line, we basically expanded the orange part on the LHS, and got the orange part on the RHS. On the second line, we realized that the RHS of the previous line has two cancelling terms (which I highlighted), which I cancelled. The third line , relates the RHS of the second line to a familiar quantity by definition.

Finally, we return to the whole fraction, and how does it look now? $$ {\frac{i{(n-1)!} + (n-i){(n-1)!}}{(n-i)!i!}} = \frac{n!}{(n-i)! i!} = \binom ni $$

and thus the result comes about.

If your still have doubts, do clarify them.

  • $\begingroup$ Thanks a lot! This is very helpful! $\endgroup$
    – aclowkay
    Sep 2, 2017 at 12:41
  • $\begingroup$ Can you clarify the orange part ? $\endgroup$
    – aclowkay
    Sep 2, 2017 at 12:42
  • 1
    $\begingroup$ The orange part? We know that there is an identity $a(b+c) =ab + ac$, where $a,b,c$ are algebraic expressions. Put $a = (n-1)!$. Put $b = n$ and $c= -i$, then $a(b+c) = (n-1)!(n-i)$, while $ab + ac = n(n-1)! - i(n-1)!$. Hence, the equality follows. $\endgroup$ Sep 2, 2017 at 12:52
  • 1
    $\begingroup$ (+1) your answer is more like someone explaining it on the board rather than just providing a solution. I can almost hear you explaining it in front of me. :) $\endgroup$
    – Krish
    Sep 2, 2017 at 14:45
  • 1
    $\begingroup$ @Krish Thank you very much. It is your encouragement that drives me. $\endgroup$ Sep 3, 2017 at 3:41

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