Q.

The product of n consecutive natural number is always divisible by

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a

nPn

b

2nCn

c

2nPn

d

n+1Pn

answer is A.

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Detailed Solution

Let n consecutive natural numbers bm+1,m+2…,m+n where m≥0P=(m+1)(m+2)…(m+n)=m!(m+1)⋯(m+n)m!n!n!= m+nCn nPn∴ P is divisible by  nPn
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