agodbole wrote:
I had a different approach to solve this problem. Let us consider the different combinations in which the letters be placed in incorrect envelopes.
We have 4 letters..
For the 1st letter, there are 3 incorrect envelopes to choose from
Similarly for the 2nd and 3rd letter there are again 3 incorrect envelopes to chose from
By this time, the 4th letter will automatically have the wrong envelope choice left.
Therefore, the number of combinations = 3*3*3*1 = 9
This is a good idea, but there are a few problems with that approach.
Let a, b, c and d represent the letters, and let A, B, C and D represent the corresponding addresses.
Step 1: Choose an envelope for letter a.
We can do this in 3 ways.
Step 2: Choose an envelope for letter b.
If, in step 1, we placed letter a in address B, then we can place b in 3 different addresses (A, C or D)
If, in step 1, we placed letter a in address C, then we can place b in 2 different addresses (A or D)
So, which is it? 3 or 2?
The problem exists for steps 3 and 4 as well.
Also, 3*3*3*1 = 27 (not 9)
Cheers,
Brent