5 letters have to be put in 5 different envelopes numbered 1 through 5 such that each of the letters go into only 1 envelope.The letter is said to be put in correct position if for example letter 1 goes into envelope 1.Now what is the probability that all letters be put into wrong envelopes?	Choose one:  A) 1/3  B) 2/3  C) 11/120  D) 11/30  E) 76/120
This Q is based on derangement, a permutation in which all elements are in the wrong position.   Number of derangements = n!(12!−13!+14!+...+((−1)n)n!)n!(12!−13!+14!+...+((−1)n)n!).. Of Course it can be derived but thats not required..   Since there are 5 letters and 5 envelopes:-   Derangements=5!(12!−13!+14!−15!)Derangements=5!(12!−13!+14!−15!)   =&gt; 120(1/2−1/6+1/24−1/120)=60−20+5−1=44.120(1/2−1/6+1/24−1/120)=60−20+5−1=44.   Total possible ways = 5! = 120.   therefore required prob, P=44/120=11/30.   ans 11/30
<result>D</result>
5 letters have to be put in 5 different envelopes numbered 1 through 5 such that each of the letters go into only 1 envelope.The letter is said to be put in correct position if for example letter 1 goes into envelope 1.Now what is the probability that all letters be put into wrong envelopes?