I will try to explain this in very simple terms:
let us consider there are four letters (L1,L2,L3,L4) with corresponding right addressed envelopes (e1,e2,e3,e4)
Now if we assume - one correct
Total possible combination = 24 (P4)
Consider => L1-e1 is right => then L2 has 2 possible combinations such as e3 or e4 (we consider L2-e3); it cannot take e2 bcoz then there will be a combination with 2 letters in right envelop. Now the obvious choice for L3 to take e4 and not e3 (same logic for L2) and thus L4 is paired bydefault with e2.....Now plz refer below sequence with 2 possibilities
Hence => [L1-e1 | L2-e3 | L3-e4 | L4-e2] OR [L1-e1 | L2-e4 | L3-e2 | L4-e3]
Now this sequence is only for L1-e1 to be the only right and all others are wrong; similarly we have 4 pairs such as (L1-e1 | L2-e2 | L3-e3 | L4-e4 ) with 2 possibilities for each
Hence Ans is 4 x 2 = 8 ways => 8/24 => 1/3
deepak123gmat wrote:Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If 4 letters are to be put into 4 envelope at random, what is the probobility 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
[spoiler]
OA - D[/spoiler]
