Question

$m$ distinct odd natural numbers and $n$ distinct even natural numbers are written at random to form a number of $(m+n)$ digits. If $n>m$, the probability two odd numbers are not together is

• A. $\frac{n!(n+1)!}{(m+n)!(n-m+1)!}$
• B. $\frac{{}^{n+1}{C}_m}{{(}^{m+n}{C}_m)}$
• C. $\frac{m!(n+1)!}{(m+n)!(n-m+1)!}$
• D. $\frac{1}{(m!{)}^2{(}^{m+n}{C}_n){(}^{n-m+1}{C}_m)}$

Answer 1

Answer: (A), (B)

$\frac{n!(n+1)!}{(m+n)!(n-m+1)!}$
$\frac{{}^{n+1}{C}_m}{{(}^{m+n}{C}_m)}$

Place $(m-1)$ even numbers between the $m$ odd numbers. There are $n-m+1$ even numbers left. These can be distributed in the $m+1$ gaps (including the extremities) in $C_{(m+1)-1}^{(n-m+1)+(m+1)-1}=C_m^{n+1}$ ways. The numbers can now be arranged in $C_m^{n+1}\times n!\times m!$
Hence probability = $\frac{C_m^{n+1}\times n!\times m!}{(m+n)!}=\frac{C_m^{n+1}}{\frac{(m+n)!}{n!\times m!}}=\frac{C_m^{n+1}}{C_m^{m+n}}=\frac{(n+1)!n!}{(n-m+1)!(m+n)!}$