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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.
i m getting 3/8??
How? I am getting 5/8
probability will be dearrangement/n!
Ok, got it.. my mistake. :(

so ans will be:

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