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Question
What is n+3C2 and n+6C2. ?
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Solution
nCr = n!/(n-r)!r!
n+3C2= (n+3)!/(n+3-2)!2! = (n+3)!/(n+1)!2=(n+3)(n+2)/2
n+6C2= (n+6)!/(n+6-2)!2!=(n+6)!/(n+4)!2=(n+6)(n+5)/2
n+3C2= (n+3)!/(n+3-2)!2! = (n+3)!/(n+1)!2=(n+3)(n+2)/2
n+6C2= (n+6)!/(n+6-2)!2!=(n+6)!/(n+4)!2=(n+6)(n+5)/2
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