Skip to content
Related Articles

Related Articles

Improve Article
GATE | GATE MOCK 2017 | Question 9
  • Difficulty Level : Expert
  • Last Updated : 06 Jan, 2017

Four different pens (1, 2, 3, 4) are to be distributed at random in four pen stands marked as 1, 2, 3, 4. What is the probability that none of the pen occupies the place corresponding to its number ?
(A) 17/24
(B) 3/8
(C) 1/2
(D) 5/8


Answer: (B)

Explanation: here total cases will be 4!= 24.
Now cases in favour as per given question –

A    B    C    D (fix pen 2 in A)
2    1    4    3
2    4    1    3
2    3    4    1 - 3 cases

Just like that fixing pen 3 and 4 at A will produce 3 cases each.
Therefore total favourable cases will be 9.

Probability = 9/24 = 3/8


Quiz of this Question

Attention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :