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