ssandeepan wrote:

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?

A) 8

B) 9

C) 10

D) 12

E) 15

I get only 9:

Envelope A can get b,c or d:

If A gets b, B can get a,c, or d. If B gets a, then c goes in D and d in C. If B gets c, then d goes in C and a goes in D

If A gets c, B can get a or d. If B gets a, then C gets d and D gets b. If B gets d, then C and D get a and b or vice versa.

If A gets d, B can get a or c. If B gets a, then C gets b and D gets C. If B gets c, then C and D get a and b or vice versa.

Perhaps it is better to count the converse: 1 way for 4 correct, 0 ways for 3 correct, 4C2 = 6 ways for 2 correct and 8 ways for only one correct ( 4C1 multiplied by 2). 24 - 1 - 6 - 8 = 9

I like this:

Think of two sets: one with a and b, the other with c and d

3 cases:

case 1: no exchange of letters between the sets- the two sets must each do the only inward swap possible (1)

case 2: one exchange of letters between the sets 2C1 x 2C1 (choose one letter for each set for the exchange, after the exchange, two sets must each do the inward swap (4)

case 3: each letter in the first set moves to the second and vice versa- the new sets can be permuted in 2 ways each (4)

4 + 4 + 1 = 9