Questions
The product of n consecutive natural number is always divisible by
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a
nPn
b
2nCn
c
2nPn
d
n+1Pn
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Correct option is A
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 nPnSimilar Questions
Let (where, [ . ] denotes greatest integer function) is equal to
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