Help me prove
$$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}.$$
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3Write down the LHS in terms of factorials and make up into one fraction – David Quinn Sep 11 '15 at 15:33
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1@DavidQuinn : I would prefer a bijective argument. ${}\qquad{}$ – Michael Hardy Sep 11 '15 at 15:36
4 Answers
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Hint: Suppose you have $n+1$ objects and you want to choose $r$ of them, but one of them is $\mathcal{Special}$.
Break it into cases: either you do pick the $\mathcal{Special}$ object or you don't. How many ways are there of doing each?
JMoravitz
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The binomial coefficient ${n}\choose{r}$ is the number of $r$-element subsets of an $n$-element set. So let's consider an $(n+1)$-element set, say the set contains $n$ pennies and one nickel (or $n$ of one kind of thing and one of a different kind of thing). How many $r$ element subsets does this set have? We can choose $r$ pennies, or $r-1$ pennies and one nickel.
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$\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-(r-1))!}=\frac{(n-(r-1))\times n!}{r!(n-(r-1))!}+\frac{r\times n!}{r!(n-(r-1))!}=\frac{(n+1)\times n!}{r!(n-(r-1))!}=\frac{(n+1)!}{r!((n+1)-r))!}$
Thus they are equivalent
Jack Pedley
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$n-(r-1) = n+1-r$. They are equivalent, as shown by several other methods. – JMoravitz Sep 11 '15 at 15:47
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You can start from the definition of the binomial coefficient $\dbinom nr$ as the coefficient of $x^r$ when you develop $(1+x)^n$, and compare the coefficients of $x^r$on both sides of the equation: $$(1+x)^{n+1}=(1+x)(1+x)^n$$ Comparing, you clearly have $\;\dbinom {n+1}r x^r=1\cdot\dbinom nrx^r+x\cdot\dbinom n{r-1}x^{r-1}.$
Bernard
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