If m rupee coins and n ten paise coins are pl | vedclass

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Question

(B) Total number of ways to arrange $m$ rupee coins and $n$ ten paise coins in a line is given by the permutation of multiset formula: $\frac{(m+n)!}{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Total number of ways to arrange $m$ rupee coins and $n$ ten paise coins in a line is given by the permutation of multiset formula: $\frac{(m+n)!}{m!n!}$.If the extreme coins are both ten paise coins,we fix two ten paise coins at the two ends. The remaining $(m+n-2)$ coins consist of $m$ rupee coins and $(n-2)$ ten paise coins.The number of ways to arrange these remaining coins is $\frac{(m+n-2)!}{m!(n-2)!}$.The required probability is the ratio of favorable arrangements to total arrangements:$P = \frac{\frac{(m+n-2)!}{m!(n-2)!}}{\frac{(m+n)!}{m!n!}}$$P = \frac{(m+n-2)!}{m!(n-2)!} 	imes \frac{m!n!}{(m+n)!}$$P = \frac{n(n-1)}{(m+n)(m+n-1)}$.

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

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