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Question

If the second term of the expansion [a113+aa1]n is 14a52, then the value of nC3nC2 is

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Solution

T2= nC1(a113)n1aa=14a52n(a)n113=14an(a)n1413=14
Comparing L.H.S. and R.H.S.
an1413=a0n=1414C314C2=14!3!11!×2!12!14!=4

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