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Bunuel I have a basic question. In this case we have 4 letters and 4 envelopes. So lets say the letters are L1-L4 and envelopes are E1-E4.. Now there are 4! combinations between the two.. I am trying to understand how we arrive at that.. If I use the logic that I select 1 letter from the 4 letters and 1 envelope from the 4 envelopes and pair them together we get 4C1 x 4C1 = 16 combinations.. Where is my thinking wrong? I understand 4! as we can select one of the 4 for the first envelope... 4x3x2x1=4!, but where is 16 missing the combinations? Thanks

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.

Re: Letter arrangements: understanding probability and combinats [#permalink]

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23 Sep 2010, 12:33

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 [but can some one give me a pure math formula?] thanks.
_________________

D Day is April 23rd, 2010 Be humble, be focused, and be calm!

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 [but can some one give me a pure math formula?] thanks.

Re: Letter arrangements: understanding probability and combinats [#permalink]

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29 Sep 2010, 10:10

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.

Bunuel 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 the 4 letters are to be put in 4 envelopes at random, what is the probability:

There are 4! = 24 possible arrangements of letters in envelopes

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

I said that for the first letter, you have a choice of 3 incorrect envelopes, for the second you have a choice of 2 incorrect envelopes, and then for the last two there is only one choice apiece. So 3*2*1*1 = 6 ways, and the probability is 6/24 = 1/4. But I'm pretty sure this is wrong, I just can't figure out the best way to do it without writing out all 24 combinations

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

Only one combination will match them all correctly, so 1/24

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

Choose the letter that will be correct - 4c1 = 4. The 3 remaining letters need to all be arranged incorrectly. There are six ways to arrange these 3 letters, and two of those arrangements will result in all three of them being wrong (I wrote out all 6 combinations to show this, I'm not sure how to do it mathematically). 4*2 = 8 ways, so the probability is 8/24 = 1/3

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

If two letters are placed correctly, then we're left with two envelopes and two letters. There are only two ways to arrange them - correctly or incorrectly. We want them to be incorrect (or else all 4 would be right), so we just need to choose the two letters that we're placing correctly. 4c2 is 6, so 6/24 or 25%

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

0%. If 3 letters are placed correctly then the 4th must also be in its correct envelope, as there is no incorrect one left for it to go into

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

Prob of one letter = 1/3, prob of no letters = 1/4, combined is (1/3) + (1/4) = (7/12). Then the probability of more than 1 is 1 - (7/12) = 5/12. I know this is wrong because my answer to (A) is wrong, but that's how I solved it

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

Since there's no way to arrange 3 correct envelopes and 1 incorrect envelope (as I showed in D), this must be the same as placing all letters correctly, or 1/24

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

Answers to follow after discussion.

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...

Re: Letter arrangements: understanding probability and combinats [#permalink]

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06 Feb 2011, 12:31

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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.

Re: Letter arrangements: understanding probability and combinats [#permalink]

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19 Feb 2011, 12:52

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Bunuel wrote:

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:

Solved using exclusion-inclusion

Bunuel wrote:

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

Re: Letter arrangements: understanding probability and combinats [#permalink]

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13 Oct 2011, 16:45

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) ?

Re: Letter arrangements: understanding probability and combinats [#permalink]

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02 Feb 2012, 17:34

[/quote]

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.[/quote]

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

Re: Letter arrangements: understanding probability and combinats [#permalink]

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20 Mar 2012, 18:12

Bunuel wrote:

rohitsb wrote:

Great post for someone like me who shivers seeing probability questions. Request a more detail explanation for question F & G.

Regards

"Problems are Purposeful Roadblocks Offering Beneficial Lessons (to) Enhance Mental Strength. Inner strength comes from struggle and endurance, not when you are free from problems."

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.

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

Re: Letter arrangements: understanding probability and combinats [#permalink]

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02 Jul 2012, 20:37

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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
_________________

Re: Letter arrangements: understanding probability and combinats [#permalink]

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03 Jul 2012, 02:04

mainhoon wrote:

Bunuel wrote:

arjunsridhar84 wrote:

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

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.

ichha148 wrote:

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

I understand that correct answer is 9/24 , however my question is should not the B. That all letters will be put into the envelope with its correct address? is opposite of no letter will be put into the envelope with its correct address?

So , should not the result be 1-1/24 = 23/24

Can some one please explain me why this is not 23/24 and when 23/24 is applicable

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.

Bunuel I have a basic question. In this case we have 4 letters and 4 envelopes. So lets say the letters are L1-L4 and envelopes are E1-E4.. Now there are 4! combinations between the two.. I am trying to understand how we arrive at that.. If I use the logic that I select 1 letter from the 4 letters and 1 envelope from the 4 envelopes and pair them together we get 4C1 x 4C1 = 16 combinations.. Where is my thinking wrong? I understand 4! as we can select one of the 4 for the first envelope... 4x3x2x1=4!, but where is 16 missing the combinations? Thanks

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
_________________

Re: Letter arrangements: understanding probability and combinats [#permalink]

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06 Jan 2013, 05:19

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)

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