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Letterarrangements: understanding probability and combinats [#permalink]
06 Oct 2009, 13:41
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00:00
A
B
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Difficulty:
(N/A)
Question Stats:
70% (01:12) correct
30% (00:00) wrong based on 22 sessions
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. _________________
Re: Letterarrangements: understanding probability and combinats [#permalink]
07 Oct 2009, 04:10
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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
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
on similar lines evaluate probability by dividing by 24 for each case
1 correct = 8 ways 2 correct = 4 ways 3 correct = 0 way ( if 3 are in right envelop 4th one will automatically go in the correct envelop) 4 correct = 1 way ( only one possible combination ) _________________
Bhushan S. If you like my post....Consider it for Kudos
Re: Letterarrangements: understanding probability and combinats [#permalink]
07 Oct 2009, 04:28
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bhushan252 wrote:
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
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
on similar lines evaluate probability by dividing by 24 for each case
1 correct = 8 ways 2 correct = 4 ways 3 correct = 0 way ( if 3 are in right envelop 4th one will automatically go in the correct envelop) 4 correct = 1 way ( only one possible combination )
You answered B. C. D. E. and one of the answers is incorrect. Try again.
A. F. and G. are left. Good luck. _________________
Re: Letterarrangements: understanding probability and combinats [#permalink]
07 Oct 2009, 07:22
1
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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:
There are \((4P4) = 4!\) ways to put the letters in envelopes
A. That no letter will be put into the envelope with its correct address?
P that at least one letter will be put in correct envelope = \(4*[(2P1)/(2P1)*(3P3)] = 4/6\) P that no letter will be put in correct envelope = \(1-4/6 = 1/3\)
B. That all letters will be put into the envelope with its correct address?
P that ALL letters will be put in correct envelope = \(1/24\)
C. That only 1 letter will be put into the envelope with its correct address?
Consider that only first letter will be put correctly and other are all incorrect. P that first letter is put correctly = \((2P1)/(2P1*3P3) = 1/6\) P that at least one letter of remaining 3 letters put correctly = \(3*[(2P1)/(2P1)*(2P2)] = 1/2\) P that NO letter of remaining 3 letters put correctly = \(1-1/2 = 1/2\) P that ONLY 1 letter put correctly = \(1/6*1/2 = 1/12\)
D. That only 2 letters will be put into the envelope with its correct address?
Consider that first and second letter will be put correctly and other are all incorrect. P that first letter is put correctly = \((2P1)/(2P1*3P3) = 1/6\) P that second letter is put correctly = \((2P1)/(2P1*2P2) = 1/2\) P that at least one letter of remaining 2 letters put correctly = \(1/2\) P that NO letter of remaining 2 letters put correctly = \(1-1/2 = 1/2\) P that ONLY 2 letter put correctly = \(1/6*1/2*1/2 = 1/24\)
E. That only 3 letters will be put into the envelope with its correct address?
P that ONLY 3 letter put correctly = \(0\), It is not possible to put only 3 letters correctly
F. That more than one letter will be put into the envelope with its correct address?
P that either 2, 3 or 4 letters correctly put = D+E+B = \(1/24+0+1/24 = 1/12\)
G. That more than two letters will be put into the envelope with its correct address?
P that either 3 or 4 letters correctly put = E+B = \(0+1/24 = 1/24\)
Note that each Q could be solved in different ways, so check your answers with the alternate solution.
Re: Letterarrangements: understanding probability and combinats [#permalink]
07 Oct 2009, 07:49
Expert's post
1
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hgp2k wrote:
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:
There are \((4P4) = 4!\) ways to put the letters in envelopes
A. That no letter will be put into the envelope with its correct address?
P that at least one letter will be put in correct envelope = \(4*[(2P1)/(2P1)*(3P3)] = 4/6\) P that no letter will be put in correct envelope = \(1-4/6 = 1/3\)
B. That all letters will be put into the envelope with its correct address?
P that ALL letters will be put in correct envelope = \(1/24\)
C. That only 1 letter will be put into the envelope with its correct address?
Consider that only first letter will be put correctly and other are all incorrect. P that first letter is put correctly = \((2P1)/(2P1*3P3) = 1/6\) P that at least one letter of remaining 3 letters put correctly = \(3*[(2P1)/(2P1)*(2P2)] = 1/2\) P that NO letter of remaining 3 letters put correctly = \(1-1/2 = 1/2\) P that ONLY 1 letter put correctly = \(1/6*1/2 = 1/12\)
D. That only 2 letters will be put into the envelope with its correct address?
Consider that first and second letter will be put correctly and other are all incorrect. P that first letter is put correctly = \((2P1)/(2P1*3P3) = 1/6\) P that second letter is put correctly = \((2P1)/(2P1*2P2) = 1/2\) P that at least one letter of remaining 2 letters put correctly = \(1/2\) P that NO letter of remaining 2 letters put correctly = \(1-1/2 = 1/2\) P that ONLY 2 letter put correctly = \(1/6*1/2*1/2 = 1/24\)
E. That only 3 letters will be put into the envelope with its correct address?
P that ONLY 3 letter put correctly = \(0\), It is not possible to put only 3 letters correctly
F. That more than one letter will be put into the envelope with its correct address?
P that either 2, 3 or 4 letters correctly put = D+E+B = \(1/24+0+1/24 = 1/12\)
G. That more than two letters will be put into the envelope with its correct address?
P that either 3 or 4 letters correctly put = E+B = \(0+1/24 = 1/24\)
Note that each Q could be solved in different ways, so check your answers with the alternate solution.
Answers to follow after discussion.
Please correct me if I am wrong.
GREAT questions Bunuel. +1 for you
That' why I said that these kind of problems are often confusing:
Total number of combinations 4!=24 - correct!
BUT most of your answers - wrong. Try again. Good practice to master. You are close to find the wright pattern.
I will post the answers after discussion, think this is the bets way. _________________
Re: Letterarrangements: understanding probability and combinats [#permalink]
08 Oct 2009, 15:18
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Great post Bunuel +1 Not sure if my answers are right cos I’m no good at these type of questions.
Total possibilities are 4! = 4x3x2x1 = 24
My methodology: (letter1 into an envelope) x (letter 2 into remaining) x … etc x (Ways to choose this) Probability 1 is correct = 1x2x1x1x4C1 = 8 Probability 2 are correct = 1x1x1x1x4C2= 6 Probability 3 are correct = 0 (Not sure about this one but I thought you can’t put only 3 in the right envelope because the last one will match the remaining letter) Probability 4 are correct 1x1x1x1x1x4C1 = 1 Probability all are wrong = 24 - (the above) = 24 – 15 = 9/24
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? 8/24 D. That only 2 letters will be put into the envelope with its correct address? 6/24 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? 1-(9/24) = 15/24 G. That more than two letters will be put into the envelope with its correct address? 17/24
Re: Letterarrangements: understanding probability and combinats [#permalink]
08 Oct 2009, 16:30
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yangsta8 wrote:
Great post Bunuel +1 Not sure if my answers are right cos I’m no good at these type of questions.
Total possibilities are 4! = 4x3x2x1 = 24
My methodology: (letter1 into an envelope) x (letter 2 into remaining) x … etc x (Ways to choose this) Probability 1 is correct = 1x2x1x1x4C1 = 8 Probability 2 are correct = 1x1x1x1x4C2= 6 Probability 3 are correct = 0 (Not sure about this one but I thought you can’t put only 3 in the right envelope because the last one will match the remaining letter) Probability 4 are correct 1x1x1x1x1x4C1 = 1 Probability all are wrong = 24 - (the above) = 24 – 15 = 9/24
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? 8/24 D. That only 2 letters will be put into the envelope with its correct address? 6/24 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? 1-(9/24) = 15/24 G. That more than two letters will be put into the envelope with its correct address? 17/24
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.
Re: Letterarrangements: understanding probability and combinats [#permalink]
08 Oct 2009, 16:53
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Bunuel wrote:
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.
Damn... I didn't read those questions properly... thanks for the correct answers This is a great post.
Re: Letterarrangements: understanding probability and combinats [#permalink]
08 Jan 2010, 15:54
yangsta8 wrote:
Great post Bunuel +1 Not sure if my answers are right cos I’m no good at these type of questions.
Total possibilities are 4! = 4x3x2x1 = 24
My methodology: (letter1 into an envelope) x (letter 2 into remaining) x … etc x (Ways to choose this) Probability 1 is correct = 1x2x1x1x4C1 = 8 Probability 2 are correct = 1x1x1x1x4C2= 6 Probability 3 are correct = 0 (Not sure about this one but I thought you can’t put only 3 in the right envelope because the last one will match the remaining letter) Probability 4 are correct 1x1x1x1x1x4C1 = 1 Probability all are wrong = 24 - (the above) = 24 – 15 = 9/24
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 .
Re: Letterarrangements: understanding probability and combinats [#permalink]
10 Jan 2010, 09:19
Expert's post
GMATMadeeasy wrote:
Bunuel wrote:
GMATMadeeasy wrote:
Part of this question :
what is the probability that all the letters are not placed in the right envelopes?
It means "no letter should be in correct envelope" or "at least one letter is in incorrect envelope " ?
Are you referring to "A. That no letter will be put into the envelope with its correct address"?
It means all 4 letters are in the envelopes with incorrect address.
Thanks. I had problem in understanding if the question is phrased in different manner. So
Quote:
all the letters are not placed in the right envelopes
means
Quote:
no letter will be put into the envelope with its correct address
?
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. _________________
Re: Letterarrangements: understanding probability and combinats [#permalink]
11 Jan 2010, 01:52
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." _________________
" 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 " Cheers !
Re: Letterarrangements: understanding probability and combinats [#permalink]
11 Jan 2010, 22:17
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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. _________________
Re: Letterarrangements: understanding probability and combinats [#permalink]
05 Mar 2010, 19:31
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.
Re: Letterarrangements: understanding probability and combinats [#permalink]
10 Mar 2010, 03:39
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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) --> \(1-\frac{1}{24}=\frac{23}{24}=0+\frac{6}{24}+\frac{8}{24}+\frac{9}{24}=\frac{23}{24}\).
Re: Letterarrangements: understanding probability and combinats [#permalink]
19 Jul 2010, 11:40
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vittarr wrote:
Hi Bunuel, could you please explain me where i was wrong for the "No letter to right enveloper" option:
I used:
P(to select 1st letter)*P(to select 1st envelope wrong) * ... * P(to select last letter)*P(to select last envelope wrong)
and i ve got 1/4: (4/4*3/4)*(3/3*2/3)*(2/2*1/2)*(1/1*1/1) = 6/24 = 1/4 (first fraction probability for letter selection, second for envelope)
this method worked for all letters in correct envelopes (4/4*1/4)*(3/3*1/3)*(2/2*1/2)*(1/1*1/1) = 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.
Re: Letterarrangements: understanding probability and combinats [#permalink]
20 Jul 2010, 11:14
Bunuel wrote:
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.
Thanks for this great post. I am still having trouble with two concepts:
1. Can you explain why we use the same choose function to establish that we have one letter in the correct envelope (4C1) as we would to determine the number of ways to pick one letter out of four out of a hat (also 4C1)? In the former we have four ways to pick a letter, and four ways to place that letter into an envelope while in the latter we simply have four ways to pick an envelope. I understand that there are four ways to correctly have only one letter/envelope pair, but I don't follow how this follows to the 4C1 notation. For example, this notation would not work if there were four letters but only two envelopes as it would if we were simply picking only one of four letters.
2. As you state above, in the case that one letter has been correctly chosen we multiply by 2 to count the number of ways of incorrectly placing the other three letters. Is there a generalization to "choose" these ways rather than counting them? For example, if we were picking only one correct pair but had 5 (or 10) possible letter/envelope pairs instead of four is there a way to find this factor without enumerating them by hand? _________________
If you find my posts useful, please award me some Kudos!
Re: Letterarrangements: understanding probability and combinats [#permalink]
29 Aug 2010, 11:17
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 _________________
Consider kudos, they are good for health
gmatclubot
Re: Letterarrangements: understanding probability and combinats
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29 Aug 2010, 11:17
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