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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. 17/18
b. 3/8
c. 1/2
d. 5/8

How will we use the formula for derangements here? Using the direct formula of !n is giving wrong answer.
asked in Numerical Ability by Loyal (3.6k points)   | 111 views
i m getting 3/8??
How? I am getting 5/8
probability will be dearrangement/n!
Ok, got it.. my mistake. :(

1 Answer

+4 votes
Best answer
Ques is asking the probability
so ans will be:

$Probability$ $ = \frac{Dearrangement}{Total\_Permutations}\\ =\frac{n!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})}{n!}\\ =1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\\ =\frac{1}{2}-\frac{1}{6}+\frac{1}{24}\\ =\frac{12-4+1}{24}\\ =\frac{9}{24} = \frac{3}{8}$
answered by Boss (9.4k points)  
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