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Question
Prove that:
\[\frac{n!}{(n - r)! r!} + \frac{n!}{(n - r + 1)! (r - 1)!} = \frac{(n + 1)!}{r! (n - r + 1)!}\]
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Solution
\[ LHS = \frac{n!}{\left( n - r \right)!r!} + \frac{n!}{\left( n - r + 1 \right)!}\]
\[ = \frac{n!}{\left( n - r \right)!r!} + \frac{n!}{(n - r + 1) [(n - r)!]}\]
\[ = \frac{n!\left( n - r + 1 \right) + n!r!}{r!\left( n - r + 1 \right) [(n - r)!]}\]
\[ = \frac{n!\left( n + 1 \right) - n!r! + n!r!}{r!\left( n - r + 1 \right)\left( n - r \right)!}\]
\[ = \frac{n!(n + 1)}{r!\left( n - r + 1 \right)\left( n - r \right)!}\]
\[ = \frac{\left( n + 1! \right)}{r!\left( n - r + 1 \right)!} = \text{RHS}\]
\[ \text{Hence proved} .\]
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