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 number of permutations of $m+n$ objects where $m

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 number of permutations of $m+n$ objects where $m$ are of one type and $n$ are of another type: $\frac{(m + n)!}{m! n!}$.If the extreme coins are 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 $\frac{(m + n - 2)!}{m! (n - 2)!}$.Therefore,the required probability is:$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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