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no its not nCr its nPr.
i.e ( n! ) / (n-r)!
since in our problem r is always 2
so the formula becomes
(n * (n-1) * (n-2)!) / (n-2)!
canceling the (n-2)! gives us
n*(n-1)
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Sherlock and Pairs
You are viewing a single comment's thread. Return to all comments →
no its not nCr its nPr.
i.e ( n! ) / (n-r)!
since in our problem r is always 2
so the formula becomes
(n * (n-1) * (n-2)!) / (n-2)!
canceling the (n-2)! gives us
n*(n-1)