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Prove that `n(n-1)(n-2) ...(n-r+1)=(n!)/((n-r)!). ` |
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Answer» We have `n(n-1)(n-2) ...(n-r+1)` ` =(n(n-1)(n-2)...(n-r+1).(n-r)!)/((n-r)!) ` [ multiplying num. and denom. by `(n-r)! `] `=(n!)/((n-r)!). ` Hence, ` n(n-1)(n-2)...(n-r+1)=(n!)/((n-r)!). ` |
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