Problem Solving

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]

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

P (Getting one correct) = 1/4
P (Getting one wrong from remaining 3) = 2/3
P (Getting one wrong from remaining 2) = 1/2
P (Getting one wrong from remaining 1) = 1

Total Probability of getting 1 correct and 3 wrong = 1/4 * 2/3 * 1/2 * 1 = 1/12

This combination is CWWW (C is correct and W is wrong).

We can have 4 combinations like this, CWWW, WCWW, WWCW, WWWC.

Thus required probability = 4*(1/12) = 1/3

This problem is using concept called "De-Arrangements". The no of de-arrangement for N distinct stuffs is
N![1/2! - 1/3! + 1/4! .... + (-1)**n 1/N!]

Total ways of arranging 4 letters in 4 places 4! = 24.
Choose one letter 4C1 ways = 4. De-Arrange remaining 3 i.e. 3! [1/2! - 1/3!] = 2.

So Probability is 4 * 2 / 24 i.e 1/3

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]

leaning on the explanations given above, I would appreciate further insight
into my question:

how do we know when to multiply events A x B, or when to take them as
arrangements - A Permutation B (A(A-1)(A-2).....into B places)

Examples:
(i) Tanya has 4L and 4E (letters and Envelops respectively).
(ii) Tanya has 4S and 5P (shirts and pair of pants respectively).

According to the explanations above total possible outcome of EL (whether
right or wrong) = 4P4 = 4x3x2x1 = 24 (why not 4x4?)

(ii), according to Lunapower (RonPurewal: Thursdays with Ron on
Manhattangmat.com), total outfits of choosing a shirt AND a pair of pants=4x5
=20..

Can someone please clarify situations regarding when to use permutations and
when to simply multiply the values...