GATE CSE
First time here? Checkout the FAQ!
x
0 votes
110 views
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)   | 110 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)  
selected by


Top Users Mar 2017
  1. rude

    4272 Points

  2. sh!va

    2994 Points

  3. Rahul Jain25

    2804 Points

  4. Kapil

    2608 Points

  5. Debashish Deka

    2254 Points

  6. 2018

    1514 Points

  7. Vignesh Sekar

    1344 Points

  8. Akriti sood

    1262 Points

  9. Bikram

    1258 Points

  10. Sanjay Sharma

    1016 Points

Monthly Topper: Rs. 500 gift card

21,454 questions
26,775 answers
60,982 comments
22,994 users