Counting 1 correct: 4C1=4, choosing which letter will be put in correct envelope. (Let's assume A letter was put in correct envelope).
Then 3 letters (BCD) and 3 envelopes (BCD) are left. # of ways to put them incorrectly is 2:
Envelopes: B-C-D
Letters: C-D-B
OR: D-B-C
So total # of ways one letter will be put into the envelope with its correct address is: 4C1*2=8. As total # of ways to distribute 4 letter in 4 envelopes is 4!=24, probability P(C=1)=8/24.

Counting 2 correct: 4C2=6, choosing which 2 letter will be put in correct envelope. (Let's assume A and B letter were put in correct envelopes).
Then 2 letters (CD) and 2 envelopes (CD) are left. # of ways to put them incorrectly is 1:
Envelopes: C-D
Letters: D-C
So total # of ways two letters will be put into the envelope with its correct address is: 4C2*1=6. As total # of ways to distribute 4 letter in 4 envelopes is 4!=24, probability P(C=2)=6/24.

Counting 3 correct: if three letters will be put in correct envelopes, then the fourth one also gets its correct envelope. So there is no way we can put exactly 3 letters in correct envelopes --> P(C=3)=0

Counting 4 correct: there are 4!=24 ways to distribute 4 letters in 4 envelopes and obviously only one is when all letters get their correct envelopes. P(C=4)=1/24.

Counting all incorrect, or 0 correct:
P(all incorrect)=1-(1 correct)-(2 correct)-(3 correct)-(4 correct)=1-8/24-6/24-0-1/24=9/24.




Opposite of "all letters in correct envelopes" (= 1-1/24 = 23/24), is: "at least one in wrong envelope" = "3 correct" (=0) + "2 correct" (=6/24) + "1 correct" (=8/24) + "0 correct, option A" (=9/24) --> \(1-\frac{1}{24}=\frac{23}{24}=0+\frac{6}{24}+\frac{8}{24}+\frac{9}{24}=\frac{23}{24}\).

Hope it's clear.
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Very good. Though there are some incorrect answers:

A.B. C. D. and E. correct. As for E: probability that only 3 letters will be put into the envelope with its correct address, is 0 because if you put 3 into the correct address envelopes, 4th one also gets the correct envelope.

F. Probability that more than one letter will be put into the envelope with its correct address is P(C=2)=6/24 plus P(C=3)=0 plus P(C=4)=1/24, which gives us P(C>1)=7/24
(You forgot to deduct P(C=1) in the way you were doing it)

G. Probability that more than two letters will be put into the envelope with its correct address is P(C=3)=0 plus P(C=4)=1/24, which gives us P(C>2)=1/24, the same probability as for P(C=4), because P(C=3)=0.

+1 to you.
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F. Probability that more than one letter will be put into the envelope with its correct address is the sum of the following probabilities:

P(C=2)=6/24, 4C2=6(choosing 2 letters for the envelopes with correct address)*1(as there is only one arrangement of two left letters to be placed incorrectly)/4!(Total number of combinations of 4 letters in 4 envelopes)=4C2*1/24=6/24;

P(C=3)=0, as when 3 letters are placed in correct envelopes the fourth one will also be placed in correct envelope, which means there won't be the case (P=0) when exactly 3 letters are placed in correct envelope;

P(C=4)=1/24, total 24 combinations from which only one is correct;

So, P(C>1)=6/24+0+1/24=7/24


G. Probability that more than two letters will be put into the envelope with its correct address is the sum of the following probabilities:

P(C=3)=0;
P(C=4)=1/24;

So, P(C>2)=0+1/24=1/24, the same probability as for P(C=4), because P(C=3)=0.
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Here's the method I used: Create the following table and start at the bottom.

# of Correct Letters
0:
1:
2:
3:
4:
------------------------
Total =

**********************************************************

Total = 4P4 = 4*3*2*1 = 24

4: Choose 4 to be correct = 4C4 = 1

3: Choose 3 to be correct * ways to make the last 1 incorrect = 4C3 * 0 = 0

2: Choose 2 to be correct * ways to make the last 2 incorrect = 4C2 * 1 = 6

1: Choose 1 to be correct * ways to make the last 3 incorrect = 4C1 * 2 = 8

0: Total - (1+0+6+8) = 9

**********************************************************

# of Correct Letters
0: 9
1: 8
2: 6
3: 0
4: 1
------------------------
Total = 24

**********************************************************

Use the table info to answer the questions.

A. That no letter will be put into the envelope with its correct address?
9/24 = (3*3)/(3*8) = 3/8

B. That all letters will be put into the envelope with its correct address?
1/24

C. That only 1 letter will be put into the envelope with its correct address?
8/24 = 8/(8*3) = 1/3

D. That only 2 letters will be put into the envelope with its correct address?
6/24 = 6/(6*4) = 1/4

E. That only 3 letters will be put into the envelope with its correct address?
0/24 = 0

F. That more than one letter will be put into the envelope with its correct address?
(6+0+1)/24 = 7/24

G. That more than two letters will be put into the envelope with its correct address?
(0+1)/24 = 1/24

Most of the times it's easier to show what the correct approach is than to explain why some approach didn't work.

But still: when you say that the probability of choosing wrong envelope for the first letter is 3/4 you are right, but then when you are saying that the probability of choosing wrong envelope for the second letter is 2/3 you are not. Because if for the first letter you chose the envelope of the second letter then when you are choosing wrong envelope for the second letter the probability would be 3/3 as there won't be correct envelope availabel for the second one (it was already used for the firs letter).

Solution for this problem is in my previous posts.

Hope it's clear.
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Please correct me if I am wrong.

GREAT questions Bunuel. +1 for you

Envelopes: 1, 2, 3, and 4;
Letters: A, B, C, and D;

1234
ABCD
BACD
BCAD
BCDA
...

You can notice that the # of ways to put 4 different letters in 4 different envelopes would be the # of permutations of 4 letters A, B, C, and D, which is 4!.

As for "4C1*4C1=16": it's not missing combinations it's just wrong. This formula counts the # of different pairs of envelope-letter.

1-A
1-B
1-C
1-D
2-A
2-B
...
4-C
4-D

Hope it's clear.
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One small additional correction in the above post :"Probability 4 are correct 1x1x1x1x1x4C1" this should be ""Probability 4 are correct 1x1x1x1x1x4C4" understand this migt sound obvious but not always .

very good explanation in fact .

I think it's more Verbal question than Quant.
"all the letters are not placed in the right envelopes" this is not correct wording, I think.

If it were: "not all letters are placed in correct envelopes", then this is the case: 0, 1, 2, or 3 letters in correct envelopes, so not all, which is 4.

"No letter will be put into the envelope with its correct address" means 0 letter in correct envelope.

Anyway GMAT won't give the question which won't be clear in this sence.
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Hi Bunuel - i didnt understand how yangsta arrived at 1*2*1*1 for all incorrect or for that matter 1*1*1*1 for one incorrect and so on... i didnt quite follow this logic.

kindly explain