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AP ICET 2015 Paper

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Question : 147 of 200

Marks: +1, -0

If 8 persons sit in a row at random, then the probability that 3 particular persons among them sit together in a given specified order is

$\frac{3}{28}$
$\frac{1}{8}$
$\frac{1}{56}$
$\frac{1}{7}$

Solution:

Total no. of ways 8 persons can be seated in a row = 8!

Treating the 3 persons sitting together as a single entity, we now have 5 + 1 i.e.6 persons.

They can be arranged in 6! ways.

Since these 3 persons sit together in a specified order, no internal rearrangement is possible.

∴ Required probability

= \frac{6!}{8!} = \frac{1}{8 \times 7} = \frac{1}{56}

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