Question)A secretary types 4 letters and then addresses the 4 corresponding envelopes. In how many ways can the secretary place the letters in the envelopes so that NO letter is placed in its correct envelope?
Solution: Let the letters be A, B, C and D; and the envelopes be a, b, c, and d
Where A goes to a, B to b, C to c, and D to d (in the ideal case)
However, we want all letters
wrongly assigned.
Let us start from A:
A can be assigned to: b or c or d
Let A be assigned to b:
Remaining letters: B, C, D and remaining envelopes: a, c, d
Case 1: B goes to a => C must go to d and D must go to c ..... (1 way)
Case 2: B goes to c => C must go to d and D must go to a ..... (1 way)
Case 3: B goes to d => C must go to a and D must go to c ..... (1 way)
Thus, when A is assigned to b, we have 3 cases
We would also have 3 cases for each of the cases when A is assigned to c, or when A is assigned to d
Thus, total ways = 3 + 3 + 3 = 9
Alternate Method: Let the letters be A, B, C and D; and the envelopes be a, b, c, and d
Where A goes to a, B to b, C to c, and D to d (in the ideal case)
If we randomly assign letters to envelopes, we can do that in 4! = 24 possible ways.
There are 4 possible cases:
1. All letters assigned correctly - 1 way
2. One letter assigned correctly -
Let the letter A be assigned correctly to a.
We are left with B, C, D and envelopes b, c, d
We need to assign these wrongly: there are 2 ways: B - c, C - d, D - b OR B - d, C - b, D - c
Thus, if B is assigned correctly to b, then too we will have 2 cases; similarly for C and D
Thus, there are a total of 8 ways when one letter is assigned correctly
3. Two letters assigned correctly -
We can select the 2 letters we want to assign correctly in 4C2 = 6 ways
The remaining 2 letters must be assigned to the other's envelope.
For example, if C and D need to be wrongly assigned, it must be that C goes to d and D to c - there is only 1 way
Thus, there are a total of 6 ways when two letters are assigned correctly
4. All letters are wrongly assigned (Note: you cannot have 3 letters assigned correctly and 1 incorrectly) -
Required number = 4! - (1 + 8 + 6) = 9 ways
Answer BNote: This can also be done by the method of derangements. However, I feel that, for the GMAT, knowing this concept is not necessary
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Sujoy Kr Datta | GMAT Q51 | CAT 99.99 | IIT (Indian Institute of Technology) alumnus
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