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

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Solution

LHS $=^{n - 1} C_{r} +^{n - 1} C_{r - 1}$
$= \frac{(n - 1)!}{(n - 1 - r)! r!} + \frac{(n - 1)!}{(n - 1 - r + 1)! (r - 1)!}$
$= \frac{(n - 1)!}{(n - r - 1)! r (r - 1)!} + \frac{(n - 1)!}{(n - r)! (r - 1)!}$
$\frac{(n - 1)!}{r (n - r - 1)! (r - 1)!} + \frac{(n - 1)!}{(n - r) (n - r - 1)! (r - 1)!}$
$\frac{(n - 1)!}{(n - r - 1)! (r - 1)!} (\frac{1}{r} + \frac{1}{n - r})$
$\frac{(n - 1)!}{(n - r - 1)! (r - 1)!} . \frac{n - r + r}{r (n - r)}$
$\frac{(n - 1)!}{(n - r - 1)! (r - 1)!} . \frac{n}{r (n - r)}$
$\frac{n!}{(n - r)! r!}$
$^{n} C_{r} =$ RHS
LHS=RHS
Hence proved.

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Similar questions

1 \leq r \leq n

{^{n} C}_{r} + {^{n} C}_{r - 1} = {^{n + 1} C}_{r}

n

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^{n} C_{r - 1} :^{n} C_{r} :^{n} C_{r + 1} = 20 : 35 : 42

^{n} C_{r - 1} +^{n + 1} C_{r - 1} +^{n + 2} C_{r - 1} + . . . . . . . +^{2 n} C_{r - 1}, =^{2 n + 1} C_{r^{2} - 132} -^{n} C_{r}

r

A = [\begin{matrix} 1 & 0 1 & 1 \end{matrix}],

I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

A^{n} = n A - (n - 1) I, n \geq 1.

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

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