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Question: If m rupee coins and n ten paise coins are placed in a line, then the probability that the extreme c...

If m rupee coins and n ten paise coins are placed in a line, then the probability that the extreme coins are ten paise coins is

A

m+nCm/nm{ } ^ { m + n } C _ { m } / n ^ { m }

B

n(n−1)(m+n)(m+n−1)\frac { n ( n - 1 ) } { ( m + n ) ( m + n - 1 ) }

C

m+nPm/mn{ } ^ { m + n } P _ { m } / m ^ { n }

D
Answer

n(n−1)(m+n)(m+n−1)\frac { n ( n - 1 ) } { ( m + n ) ( m + n - 1 ) }

Explanation

Solution

mm rupee coins and nn ten paise coins can be placed in a line in (m+n)!m!n!\frac { ( m + n ) ! } { m ! n ! } ways.

If the extreme coins are ten paise coins, then the remaining n−2n - 2 ten paise coins and mm one rupee coins can be arragned in a line in (m+n−2)!m!(n−2)!\frac { ( m + n - 2 ) ! } { m ! ( n - 2 ) ! } ways.

Hence the required probability

=(m+n−2)!m!(n−2)!(m+n)!m!n!=n(n−1)(m+n)(m+n−1)= \frac { \frac { ( m + n - 2 ) ! } { m ! ( n - 2 ) ! } } { \frac { ( m + n ) ! } { m ! n ! } } = \frac { n ( n - 1 ) } { ( m + n ) ( m + n - 1 ) }