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I am reading Elementary Number Theory 7th Edition from David Burton. (See here.) On page 8 he sets about to prove Pascal's Rule

$$\binom{n}{k} + \binom{n}{k - 1} = \binom{n+1}{k}$$

But then he says

Its proof consists of multiplying the identity

$$\frac{1}{k} + \frac{1}{n-k+1}=\frac{n+1}{k(n-k+1)}\tag{1}$$

by

$$\frac{n!}{(k-1)!(n-k)!}\tag{2}$$

to obtain $$ \frac{n!}{k(k-1)!(n-k)!}+\frac{n!}{(k-1)!(n-k+1)(n-k)!} =\frac{(n+1)n!}{k(k-1)!(n-k+1)(n-k)!} $$

which can then be rearranged into the factorial version of Pascal's Rule. Fine, but where did the "identity" expressions and the expression to multiply it by ($(1)$ and $(2)$) come from? Also, how is this a proof?

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    As we can see , we could just divide by $\frac{n!}{(k-1)!(n-k)!}$ our first statement. And after this we will get (1). – 2017-01-12
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    Chicken or the egg? So then, where then does $\frac{n!}{(k-1)(n-k)!}$ come from? – 2017-01-12
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    they just miltiply by it. Some of "magic" constant which will help in task. – 2017-01-12
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    Because of it i want to say that this proof isn't pretty. – 2017-01-12
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    There must be a reason for the choices of (1) and (2). Otherwise, it's not a proof. – 2017-01-12
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    That's a proof because of logical. We have right equation (1) and we get from it (2). $a \to b$ , this implication give us the rule : if a s true , then b also true. – 2017-01-12

1 Answers 1

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$\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k} |\cdot \frac{(k-1)!(n-k)!}{n!}$, so $\frac{1}{k} + \frac{1}{n-k+1} = \frac{n+1}{k(n-k+1)}$ which is true, so the statement is true.

But this looks awful. It have simple combinatorical proof:

$\binom{n+1}{k}$ - number of ways to select $k$ objects from $n+1$. Suppose we select one object , now we could get others with $\binom{n}{k-1}$ ways. But we also could not to get the first object and so there is $\binom{n}{k}$ ways to select others.