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goes into a wrong envelope?

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goes into a wrong envelope?

by sanju09 » Tue Apr 15, 2014 4:11 am
There are four letters A, B, C, and D that have to go into 4 envelopes addressed to a, b, c, and d respectively. In how many ways can the four letters be put in the 4 envelopes such that every letter goes into a wrong envelope?
A. 20
B. 12
C. 9
D. 6
E. 4

[spoiler]OA C[/spoiler]


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by GMATGuruNY » Tue Apr 15, 2014 6:15 am
sanju09 wrote:There are four letters A, B, C, and D that have to go into 4 envelopes addressed to a, b, c, and d respectively. In how many ways can the four letters be put in the 4 envelopes such that every letter goes into a wrong envelope?
A. 20
B. 12
C. 9
D. 6
E. 4

[spoiler]OA C[/spoiler]

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Let the correct ordering be ABCD.
Strategy:
Write out the possible arrangements for ONE CASE.
Use this information to determine the number of possible arrangements for the REMAINING CASES.

Case 1: A in the second position
The following arrangements are viable:
BADC
CADB
DABC
Total options = 3.

Implication:
When A is in the 3rd position, there will be 3 more options.
When A is in the 4th position, there will be 3 more options.

Total options = 3+3+3 = 9.

The correct answer is C.
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by GMATGuruNY » Tue Apr 15, 2014 6:19 am
Formula approach:

A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):
Number of possible derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Thus, given 4 letters:
Number of possible derangements = 4! (1/2! - 1/3! + 1/4!) = 12-4+1 = 9.
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