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I didn't really understand this proof to begin with, and I really don't understand what this question is asking. Do I just let n equal some number and then try to prove it from there, how do I begin the proof?

Kristen M. Day
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    Related: http://math.stackexchange.com/questions/20475/proving-pascals-rule-n-choose-r-n-1-choose-r-1n-1-choose –  Feb 23 '16 at 04:17
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    Have you ever work with the Pascal's triangle? – Ángel Mario Gallegos Feb 23 '16 at 04:19
  • Didn't understand what proof? Did you try some method and get stuck? Or perhaps are confused about the notation? – pancini Feb 23 '16 at 04:26
  • What's your definition of $\binom{n}{k}$? – Brian Tung Feb 23 '16 at 05:53
  • The question asks you to verify that the formula holds for particular values of $n$ and $k$. For instance, if $n = 5$ and $k = 3$, you must show that $\binom{5}{3} = \binom{4}{2} + \binom{4}{3}$. You can prove Pascal's formula $$\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$$ using an algebraic argument or a combinatorial argument such as that given below by Laars Helenius. – N. F. Taussig Feb 23 '16 at 12:23

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Clearly the left hand side counts the number of possible sets of size $k$ chosen from $n$ objects.

Now select one of the $n$ objects as special. Each set of size $k$ either contains our special object or it doesn't. There are $\binom{n-1}{k-1}$ sets of size $k$ that include our special object and $\binom{n-1}{k}$ that do not.

Laars Helenius
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