KVPY SX SB Exam 2013 Question Paper

Question : 89 of 120

Marks: +1, -0

A box contains coupons labelled 1,2,...,100. Five coupons are picked at random one after another without replacement. Let the numbers on the coupons be

x_1,x_2,...,x_5

. What is the probability that

x_1 > x _2 > x_3

and

x_3 < x_4 < x_5

$1/120$
$1/60$
$1/20$
$1/10$

Validate

Solution:

{^100C_5[1×2(3)]}/{^100C_5×5!}=1/20

Suppose 1, 2, 3, 4, 5 are selected coupons

Place of 1 is fixed
Total arrangements of 5 is 2 and

arrangements of 2, 3, 4, are

\table 2,3,1,4,5;4,2,1,3,5;4,3,1,2,5\}3\ways

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