Question

A number is selected at random then the probability such that no two identical digits appear together.

• A. $\frac{37}{84}$
• B. $\frac{43}{84}$
• C. $\frac{17}{84}$
• D. $\frac{23}{84}$

Answer 1

The correct option is A $\frac{37}{84}$

There are $3$ pairs and $2$ singles.

Total ways of forming 8 digit number $=\frac{8!}{2!2!2!}=5040$

The simplest way to find ways such that no two identical digits appear together is $Inclusion/Exclusion$

$=$ $Allways-wayswithatleast1pairtogether+atleast2pairstogether-all3pairstogether$

$=\frac{8!}{2!2!2!}-(\frac{3}{1})\frac{7!}{2!2!}+(\frac{3}{2})\frac{6!}{2!}-5!$

$=5040-3\times 1260+3\times 360-120$

$=5040-3780+1080-120=2220$

$\therefore probability=\frac{2220}{5040}=37/84$

Hence, $\left(A\right)$