Question

There are $4$ shillings and $3$ half-crowns placed at random in a line$:$ show that the chance of the extreme coins being both half$-$ crowns is $\frac{1}{7}$. Generalize this result in the case of m shillings and n half$-$ crowns.

Answer 1

The number of ways in which the coins can be placed is $\frac{7!}{4!3!}$ or $35$. And there are only $5$ different ways of placing the coins so that the extreme places are occupied by half crowns

$\therefore$The required chance $=\frac{5}{35}=\frac{1}{7}$

In general case, the total no. of possible arrangements is $\frac{(m+n)!}{m!n!}$, only $\frac{(m+n-2)!}{m!(n-2)!}$ of which are favourable, and thus the chance is;-

$\Rightarrow \frac{(m+n-2)!m!n!}{m!(n-2)!(m+n)!}$

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