I am trying to prove Pascal's Rule algebraically but I'm stuck on simplifying the numerator. This is the last step that I have, but I'm not sure where to go from here
$$=\frac{\left[(k-1)(n-k)!+k(n-1-k)!\right]\times(n-1)!}{k!(n-k)!}$$
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1You can find it here. http://en.wikipedia.org/wiki/Pascal's_rule#Algebraic_proof – OR. Aug 06 '13 at 15:33
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Algebraic proof can also be found in answers to this question. – Martin Sleziak Nov 06 '16 at 08:11
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I will assume that you are trying to show that $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$, where $n$ and $k$ are each $\ge 1$. Expressing the right-hand side in terms of factorials, we get
$$\frac{(n-1)!}{k!(n-k-1)!}+\frac{(n-1)!}{(k-1)!(n-k)!}.$$
We want to bring the above expression to a common denominator. So we multiply top and bottom of the first term by $n-k$, and top and bottom of the second term by $k$. We get
$$\frac{(n-1)!(n-k)+(n-1)!k}{k!(n-k)!}.$$
The bottom now looks nice. The two terms on top have a common factor of $(n-1)!$. So the top can be rewritten as
$$(n-1)![(n-k)+k)],$$
which is $(n-1)!n$, that is, $n!$. Thus we end up with $\frac{n!}{k!(n-k)!}$, which is $\binom{n}{k}$.
André Nicolas
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2@Susan You can exploit the factorial $$k! = \prod_{j=1}^k j$$ So $(n-k-1)! \cdot (n-k) = (n-k)!$ and $(k-1)! \cdot k = k!$ – AlexR Aug 06 '13 at 15:51
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@Susan: We could multiply top and bottom of the first term as you suggest, and do something similar with the second term. We then end up with a common denominator that is much larger than necessary, but that's OK. We also end up with a more complicated expression on top. That can be simplified, and after simplification we can do a lot of cancelling of common stuff from top and bottom. It is just a lot messier. Also, I had an eye on the ultimate answer, which has a $k!(n-k)!$ at the bottom. So that's a nice common denominator to aim for, and it is achievable. – André Nicolas Aug 06 '13 at 16:03