combination

[armaan700+]

Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

A) 1/24
B) 1/8
C) 1/4
D) 1/3
E) 3/8

[thephoenix]

Total # of ways of choosing envelopes=4!=24.
Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct)

ABCD(envelopes)
ACDB(letters)
ADBC(letters)
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8

P(C=1)=8/24=1/3