A certain company assigns 5 rooms to each of its 5 employees in the marketing department. Next month, Company plans to change these employees rooms in such a way that no employee get the same room. In how many ways company can assign rooms to these employees.
a) 5
b) 8
c) 44
d) 60
e) 120
The OA is the option C.
Experts, how can I solve this PS question fast? Can you show me the best approach to solve this PS question? Thanks in advanced.
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A certain company assign 5 rooms to each of its 5 employees
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The problem above is the same as the following:M7MBA wrote:A certain company assigns 5 rooms to each of its 5 employees in the marketing department. Next month, Company plans to change these employees rooms in such a way that no employee get the same room. In how many ways company can assign rooms to these employees.
a) 5
b) 8
c) 44
d) 60
e) 120
If the correct ordering for five letters is ABCDE, how many ways can the letters be arranged so that no letter is in the correct position?
A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):
Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!).
Given 5 letters A, B, C, D and E:
Total number of derangements = 5! (1/2! - 1/3! + 1/4! - 1/5!) = 60-20+5-1 = 44.
The correct answer is C.
The problem above seems too complex for the GMAT.
Here are two similar problems that can be solved efficiently without use of the formula above:
https://www.beatthegmat.com/probability-t55896.html
https://www.beatthegmat.com/probability-t121078.html
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This is called a derangement problem, and the derangement of n items is denoted by !n where !n is defined by !0 = 1 and the recursive formula (for n ≥ 1):M7MBA wrote:A certain company assigns 5 rooms to each of its 5 employees in the marketing department. Next month, Company plans to change these employees rooms in such a way that no employee get the same room. In how many ways company can assign rooms to these employees.
a) 5
b) 8
c) 44
d) 60
e) 120
!n = n(!(n - 1) + (-1)^n
We want to calculate !5. But we need !1, !2, !3, and !4 before we can get to !5.
!1 = 1(!0) + (-1)^1 = 1(1) + (-1) = 0
!2 = 2(!1) + (-1)^2 = 2(0) + 1 = 1
!3 = 3(!2) + (-1)^3 = 3(1) + (-1) = 2
!4 = 4(!3) + (-1)^4 = 4(2) + 1 = 9
!5 = 5(!4) + (-1)^5 = 5(9) + (-1) = 44
Answer: C
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