Letter arrangements: understanding probability and combinats

Question

There was a topic with problem:

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 in 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

And seems that it was confusing for many. GMAT often has similar questions, so find below the problems to master yourself in them.

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 in 4 envelopes at random, what is the probability:

A. That no letter will be put into the envelope with its correct address?

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

C. That only 1 letter will be put into the envelope with its correct address?

D. That only 2 letters will be put into the envelope with its correct address?

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

F. That more than one letter will be put into the envelope with its correct address?

G. That more than two letters will be put into the envelope with its correct address?

Note that each Q could be solved in different ways, so check your answers with the alternate solution.

Answers to follow after discussion.

Bunuel · Answer

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.

mainhoon · Answer

Thanks for the explanation. Indeed it is 4!, appreciate it.

Posted from my mobile device

gurpreetsingh · Answer

Great Thread!! I m sure after reading this I won't lose marks in similar questions. Thanks Bunnel

MBAwannabe10 · Answer

nice question: 
here is my approach: can someone help me to know where i got wrong with a?
no letter: 1st one 3/4 2nd one : since first one is already wrong, so there are actually 1/2 of getting right, then again don't count 1st letter, so it's 1/1 => 3/8

meanwhile: 
b: 1/4*1/3*1/2 = 1/24
c: 1/4*2/3*1/2 = 1/12 then there are four ways of doing this --->1/12*4 = 1/3
d: 1/4*1/3*1/2 = 1/24 then since two letters, grouping them give you 3 sets---> ways of doing 3! =>6 
so 1/24*6 = 1/4 
e: it's just not possible? so 0  
thanks.

Bunuel · Answer

Solutions are on the 1st page. For example see: letter-arrangements-understanding-probability-and-combinats-84912.html#p697651 letter-arrangements-understanding-probability-and-combinats-84912.html#p672926 letter-arrangements-understanding-probability-and-combinats-84912.html#p637147

TehJay · Answer

I'm going to post my answers before reading any of the discussion so I can review my work and see what I did wrong.

EDIT: Looks like I did alright, but I still don't know how to do (A) mathematically (without knowing the answers to the other ones first). How can you solve this just on its own? You could do it with conditional probability, but isn't that out of scope for the GMAT? It's been almost a year since I did the probability actuary exam...

cabelk · Answer

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

dimitri92 · Answer

good question and initiative. well, here's how i solved the question.

let's assume the correct arrangement of envelopes is  ABCD 
 total cases = 4! = 24

if 1st & 2nd are already chosen, 3rd & 4th will have only 1 choice to either select the correct env or the incorrect one.

Option(A)   
possible cases =   3 * 3 * 1 * 1  = 9 
1st env = 3 (all incorrect)
2nd env = 3 (all incorrect)
3rd & 4th = only 1 incorrect will be left for each.

P(A) =  \frac{9}{24}

Option (B)   
only one case where all are correctly placed i.e. ABCD
 P(B)  =  \frac{1}{24}

Option (C)   
let’s consider case where 1st is placed correctly.
possible cases =   1 * 2 * 1 * 1 * 4C1  = 8 
1st env = 1 (correct)
2nd env = 2 (2 incorrect out of possible 3)
 4C1  = choosing 1 correct out of 4. all four (A,B,C,D) will have a chance to have correct address, so 4.

P(C)  =  \frac{8}{24}

Option (D)   
let’s consider case where 1st & 2nd are placed correctly.
possible cases =   1 * 1 * 1 * 1 * 4C2  = 6 
 4C2  = choosing 2 correct out of 4.

P(D)  =  \frac{6}{24}

Option (E)   
if all 3 are placed correctly, the 4th automatically will be placed correct. So there are no cases where only 3 are placed correctly.

P(E)  =  \frac{0}{24}

Option (F)   
P(2)+P(3)+P(4) =  \frac{6}{24} + 0 + \frac{1}{24} = \frac{7}{24}

Option (G)   
P(3)+P(4) =  \frac{1}{24}

fluke · Answer

Solved using exclusion-inclusion

\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!} = \frac{9}{24}

\frac{1}{24}

Total number of ways 3 pairs can be incorrect
 3!*(\frac{1}{2!}-\frac{1}{3!})=2

Total number of ways for 1 pair correct and 3 pairs incorrect
 C^4_1*2 = 8

P = \frac{8}{24}=\frac{1}{3}

Total number of ways 2 pairs can be incorrect:
 2!*\frac{1}{2!} = 1

Total number of ways for 2 pair correct and 2 pairs incorrect
 C^4_2*1 = 6

P = \frac{6}{24}=\frac{1}{4}

0

P(Exactly two)+P(Exactly three)+P(Exactly 4)

P(Exactly three)+P(Exactly 4)

zura · Answer

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

can you elaborate it algebraically ?

Unprez · Answer

Can you go more in detail how you get A... I'm not getting this :s ...I am thinking for A the prob is of getting right is 4/24? , how is the prob for B-D being done? Shouldn't it be the same so 4(4/24) ?

Lstadt · Answer

We know that probabilities are defined as (Favored outcomes)/(Total outcomes)

With four letters, we know that the way to arrange them is 4! so Total outcomes = 24.

Now the tricky part is how to determine favored outcomes. How does one determine that? especially for A.

It would be great if Bunuel or someone could go through and give us the official answers.

Bunuel · Answer

Solutions are on the 1st page. For example see: letter-arrangements-understanding-probability-and-combinats-84912.html#p697651 letter-arrangements-understanding-probability-and-combinats-84912.html#p672926 letter-arrangements-understanding-probability-and-combinats-84912.html#p637147

Lstadt · Answer

Thank you Bunuel. Great as always.

docabuzar · Answer

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.

Just to Add some more scenarios:   
Pls correct if required!

Probability of at least 2 correct 
P(C<=2)= P(c1)+P(C2) = 14/24

Probability of at least 3 correct 
P(C<=3)= P(C1)+P(C2) + P(C3) = 14/24

Probability of at least 2 in-correct 
P(inC<=2) = P(inC2) + P(inC1)
 = P(C2) + P(C3) = 6/24 + 0 = 6/24 
 So P(inC<=2) = P(C2)

Probability of at least 3 in-correct 
P(inC<=3) = P(inC3) + P(inC2) + P(inC1)
 = P(C1) + P(C2) + P(C3) = 8/24+ 6/24 + 0 = 14/24

Dont forget Kudos pls !!

whan311 · Answer

Excellent explanation Bunuel. I used 1-x method for F and G and got the same result.

nishtil · Answer

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 in 4 envelopes at random, what is the probability?

A. 
The probability that 1st letter will not be put in correct address is 3/4
The probability that 2nd letter will not be put in correct address is 2/3
The probability that 3rd letter will not be put in correct address is ½
The probability that 4th letter will not be put in correct address is 1
Combine probability is ¾*2/3*1/2*1 = ¼
B. 
The probability that 1st letter will be put in the correct address is ¼
The probability that 2nd letter will be put in the correct address is 1/3
The probability that 3rd letter will be put in the correct address is ½
The probability that 4th letter will be put in the correct address is 1
Combine probability = ¼*1/3*1/2*1 = 1/24
C. 
The probability that 1st letter in correct address is ¼
The probability that 2nd letter not in correct address = as there are 2 wrong and 1 right address 2/3
The probability that 3rd letter not in correct address = as there are 1right and 1 wrong address, hence ½

The probability that last letter in wrong address is 1

The combine probability of 1st letter in right address and remaining letters in wrong is = ¼*2/3*1/2 = 1/12

As any one letter can be in right address we have to multiply the above probability by 4

1/12 * 4 = 1/3

Hence, probability that exactly one letter in right address is 1/3
D. 
Probability that the first letter in correct address is ¼

Probability that the second letter in correct address is 1/3

Probability that third letter in wrong address is ½

Probability that last letter in wrong address is 1

Combine probability = 1/24

There can be 6 combinations of 2 right and 2 wrong addresses

Hence probability that exactly 2 letters in right address is 1/4
E. 
Exactly 3 letters will be put in correct address means all in correct address.
If 3 letters are in correct address ,then obviously last letter will go in correct address

probability is 0

F. 
P ( more than one letter in right envelope correct ) = 1- P( exactly 1 letter in right envelope correct)
 = 1- 1/3 = 2/3
G. 
P ( more than 2 letters in right envelope correct ) = 1- p (exactly 2 letters in right envelope )
 = 1- 1/8 = 7/8

Looking at the answer by other members,I figure out I made lot of mistakes........ I am not able to figure out where I am going wrong... Please help me out with my approach

nishtil · Answer

I think we can understand it as 4 slots ( 4 envelope) available for 4 letters.

any of the 4 can go in envelope 1 so 4 ways

any of the remaining 3 can go in 2nd envelope so 3 ways

any of the remaining 2 can go in 3rd envelope so 2 ways

any of the remaining 1 can go in 1 envelope so 1 way.

using fundamental counting principle total number of ways are 4*3*2*1 = 24

sagarsir · Answer

A. That no letter will be put into the envelope with its correct address?  9/24

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?   1/3

D. That only 2 letters will be put into the envelope with its correct address?   1/4

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

F. That more than one letter will be put into the envelope with its correct address?  7/24

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

These are my answers and I I guess I am correct.

Ex. D. id clearly 1/4 because 4C2* 1 /24 = 1/4. Explanation: 2 letters to go in their correct envelopes can be selected (not permutated!!!!) as 4C2 and for each such combinations, the remaining 2 can go into wrong places in 1 manner (both crossing their positons)

KUDOS IF U LIKE

up4gmat · Answer

Thanks Bunuel for posting this!!

Letter arrangements: understanding probability and combinats

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