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

\frac{37}{84}

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\frac{43}{84}

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\frac{17}{84}

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\frac{23}{84}

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Solution

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 $I n c l u s i o n / E x c l u s i o n$
$=$ $A l l w a y s - w a y s w i t h a t l e a s t 1 p a i r t o g e t h e r + a t l e a s t 2 p a i r s t o g e t h e r - a l l 3 p a i r s t o g e t h e r$
$= \frac{8!}{2! 2! 2!} - (\frac{3}{1}) \frac{7!}{2! 2!} + (\frac{3}{2}) \frac{6!}{2!} - 5!$
$= 5040 - 3 \times 1260 + 3 \times 360 - 120$
$= 5040 - 3780 + 1080 - 120 = 2220$
$∴ p r o b a b i l i t y = \frac{2220}{5040} = 37 / 84$
Hence, $(A)$

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