There are 6 letters and 6 self addressed envelopes.What is the probability that atleast 1 is placed correctly??

If we let A be the event that letter A is in the correct envelope, and 
similarly B is event that letter B is in the correct envelope, then

P(A) = 1/5  

P(A and B) = 1/5 x 1/4

Now use the inclusion, exclusion formula to get the probability that A or 
B or C .... or E are correctly placed.

P(A or B or C .... or E) = P(A) + P(B) + P(C) + P(D) + P(E)
                         - P(A and B) - P(B and C) -....
                         + P(A and B and C) + P(B and C and D) + ....
                         - P(A and B and C and D) - P(...) -......
                         + P(A and B and C and D and E)

                         = 5 x (1/5)
                           - 5C2 x (1/5)(1/4)
                           + 5C3 x (1/5)(1/4)(1/3)
                           - 5C4 x (1/5)(1/4)(1/3)(1/2)
                           + (1/5)(1/4)(1/3)(1/2)(1/1)

      5x4    1     5x4x3      1      5x4x3x2      1          1
 1 - ---- x ---- + ------ x ----- - --------- x ------- + --------
      1x2   5x4    1x2x3    5x4x3    1x2x3x4    5x4x3x2   5x4x3x2x1

      = 1 - 1/2! + 1/3! - 1/4! + 1/5!


This is the probability that at least one letter is correctly placed.

Approved

Go from line 16 to line 20 directly as the division is not clearly visible.