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Question

if (n+1)!=12(n-1)! find the value of n (permutation and combination)

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Solution

n+1!=12n-1!n+1nn-1!=12n-1!n+1n=12n2+n=12n2+n-12=0n2+4n-3n-12=0nn+4-3n+4=0n+4n-3=0n=-4 and 3Discard the negative value. Thus,n=3

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