Question

If $m$ one 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. $m+n_{C_m}$
• B. $\frac{n(n-1)}{(m+n)(m+n-1)}$
• C. $m{+}^2{P}_m$
• D. $m{+}^n{P}_n$

Answer 1

Answer: (B)

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

Let us fix two ten paisa coins at the extremes. We are left with $m$ one rupee coins and $(n-2)$ ten paisa coins.
These can be arranged in: $\frac{(m+n-2)!}{m!\times (n-2)!}$ ways.
Hence, required probability = $\frac{\frac{(m+n-2)!}{m!\times (n-2)!}}{\frac{(m+n)!}{m!\times n!}}=\frac{n\times (n-1)}{(m+n)\times (m+n-1)}$