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I have tried using the principle of mathematical induction. The base case is simple, but I am having trouble with the induction step. Any help would be great, thanks !

Lgate8
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    can you quickly type up your induction step so we can see where you get stuck? – Calvin Khor Sep 25 '15 at 00:01
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    A store has $n$ different chocolate doughnuts and $m-n$ different sprinkles doughnuts. In how many ways can we select a total of $n$ doughnuts? Note that the formula uses the convention that $\binom{a}{b}=0$ if $b\gt a$. – André Nicolas Sep 25 '15 at 00:02
  • What do you think is the base case? – Thomas Andrews Sep 25 '15 at 00:04
  • The case when $n=1$ – Lgate8 Sep 25 '15 at 00:05
  • assuming the statement holds for $n$, I want to prove this implies it also holds for $n+1$ – Lgate8 Sep 25 '15 at 00:06
  • https://en.wikipedia.org/wiki/Vandermonde's_identity – Jack D'Aurizio Sep 25 '15 at 00:12
  • Thanks for the link! Is there maybe a way still to do it with induction? I tried to use the induction hypothesis as follows; $$\sum_{k=0}^{n+1} {n+1 \choose k}{m-n-1 \choose n+1-k} =$$ $$\sum_{k=0}^{n} {n+1 \choose k}{m-n-1 \choose n+1-k} + 1$$

    I think I just am on the wrong track. I want to somehow use the fact that $\sum_{k=0}^n {n \choose k}{m-n \choose n-k} = {m \choose n}$, I'm just not sure how.

     – Lgate8 Sep 25 '15 at 00:18
  • It would be pretty difficult to prove this by induction because this equality involves two variables $m$ and $n.$ I think you are using induction on $n,$ but that leaves out $m.$ So using your method, you would be proving that for a constant $m,$ the equality holds. – quasicoherent_drunk Sep 25 '15 at 02:51
  • http://math.stackexchange.com/questions/219928/inductive-proof-for-vandermondes-identity provides a nice induction proof. – Vizuna Sep 25 '15 at 03:23

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Although a MI method may be used, a Combinatorial argument is much more simple. We assume that we want to select n distinct objects of a total of m objects. We can always select them directly and get RHS.Alternatively, we can separate the m objects into two groups, one of n and the other of m-n.

Now assume we take 0 from the group with n, we must choose n from the group with m-n. If we take 1 from the group with n, we must Take (n-1) from the group with (m-n). We continue repeating this & sum up the result, and we get RHS.

TheFrailOne
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Another way of proving this is by comparing the coefficients of $x$ in

$$(1+x)^m = (1+x)^n(1+x)^{m-n}$$

i.e. $$\sum_{k=0}^m \binom{m}{k} x^k = \sum_{l=0}^n \binom{n}{a}x^l\sum_{l=0}^{m-n} \binom{m-n}{l}x^l$$

Where the product of two series $∑a_nx^n, ∑b_nx^n$ has coefficients $c_n = ∑_{k+l=n}a_kb_{n-l}$.

Calvin Khor
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