Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 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

Let the 4 letters be A, B, C and D.
Total ways to arrange the 4 letters = 4! = 24.
Let the correct ordering of the 4 letters be ABCD.

Write out the ways that ONLY A can be put in the correct position:
ACDB
ADBC
Total ways = 2.

Using the same reasoning, there will be 2 ways that ONLY B can be put in the correct position, 2 ways that ONLY C can be put in the correct position, and 2 ways that ONLY D can be put in the correct position.
Thus, the total number of ways to put EXACTLY 1 letter in the correct position = 2+2+2+2 = 8.

Thus:
P(exactly 1 letter is put in the correct position) = 8/24 = 1/3.

The correct answer is D.

Step 1: 4 ways to select the letter that goes into the right envelope (either A or B or C or D)

Step 2: The number of ways to place other 3 letters wrongly

(Assumed) if A goes in correct envelope
then B has 2 options (Leaving it's own envelope as only one letter should go into correct envelope)
and C has one option left (it can't go into Its own envelope and can't leave D's envelope unoccupied if its left unoccupied till previous step (Important step)
and D also has one option left

Therefore total Favorable cases = 4 x 2 x 1 = 8

Total ways to keep the letters randomly into 4 envelopes = 4! = 24

Probability = 8/24 = [spoiler]1/3 ANSWER[/spoiler]

See Img attached.

McDesty,

Your line of reasoning does not seem valid.
Please consider the problem below:

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that NO 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

OA: E

If we apply your line of reasoning to this problem, (FFFF)/(all possible cases) = 1/12, but the correct answer is [spoiler]3/8[/spoiler].

Here's one solution:

Let the correct ordering of the 4 letters be ABCD.
Number of ways to arrange the 4 letters = 4! = 24.

Since A cannot occupy the first position, there are 3 options for A:
_ A _ _
_ _ A _
_ _ _ A

Case 1: _ A _ _
Since all of the remaining letters must be incorrectly placed, we get the following options:
BADC
CADB
DABC
3 ways.

Case 2: _ _ A _
Using the reasoning above, Case 2 will yield another 3 ways to incorrectly place all of the letters.

Case 3: _ _ _ A
Using the reasoning above, Case 3 will yield another 3 ways to incorrectly place all of the letters.

Thus, the total number of ways to incorrectly place all of the letters = 3+3+3 = 9.
Thus, P(no letter is correctly placed) = 9/24 = 3/8.

The correct answer is E.

I liked your handwritten, colorful answer, McDesty, but I agree with Mitch - you don't seem to have the right sample space, as you've neglected a few possible arrangements. For instance, FFFF can be arranged in more ways than 1. Suppose the correct arrangement should be 1234. We could have four misaddressed envelopes in the following ways:

2341
2413
2143
3412
3142
3421
4123
4321
4312

Ordering is a pain, though - it seems to be the first thing I forget every time I jump up a level of difficulty in my study of probability.

Thank you Mitch and the Veritas prep instructor.(I clicked the thank you icon as well)
I will definitely leave that up so someone else doesn't repeat the same faulty line of reasoning.