Prove that $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ .

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Solution

Now,
$^{n} C_{r} +^{n} C_{r - 1}$
$= \frac{n!}{r! (n - r)!}$ $+ \frac{n!}{(r - 1)! (n - r + 1)!}$ [ Using the definition of $^{n} C_{r}$ ]
$= \frac{n!}{(n - r)! (r - 1)!} [\frac{1}{r}$ $+ \frac{1}{(n - r + 1)}]$
$= \frac{n!}{(n - r)! (r - 1)!} [\frac{(n + 1)}{r (n + 1 - r)}]$
$= \frac{(n + 1)!}{r! (n + 1 - r)!}$
$=^{n + 1} C_{r}$ .

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^{n} C_{r - 1} +^{n} C_{r} =^{n + 1} C_{r}

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