Five persons A, B,C, D and. E ate in queue of a shop. The probability that A and E are always together, is

Five persons A, B,C, D and. E ate in queue of a shop. The probability that A and E are always together, is

A
$\frac{1}{4}$
B
$\frac{2}{3}$
C
$\frac{2}{5}$
D
$\frac{3}{5}$

Solution:

Total number oi ways' = 5!
and favorable numbers of ways = 2 .4!
.'. Required probability $= \frac{2 \cdot 4!}{5!} = \frac{2}{5}$

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