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Re: Permutation Problem [#permalink]
why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?
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Re: Permutation Problem [#permalink]
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gmatJP wrote:
why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?

there are 8 nos in total therefore there are 8! ways to arrange them....
however A is used two times and B three times.....so A can be arranged within itself 2! and similarly B 3! times.
since all B and A are the same the times these can be arranged within themselves are the same combination ...
so distinct combinations would be 8!/2!3!...
and ofcourse the restr is c on right of d... it is clear that in half cases it is possible so divide the soln above by 2

Originally posted by chetan2u on 24 Dec 2009, 22:00.
Last edited by chetan2u on 25 Dec 2009, 01:10, edited 1 time in total.
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Re: Permutation Problem [#permalink]
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Bunuel wrote:
tania wrote:
In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D?
A.1680
B.2160
C.2520
D.3240
E.3360

Can someone explain how I should approach to solve the above problem?


We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\).

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\)

Answer: A.


Can you please explain the logic or how you could deduce quickly the following:-

Now, in half of these cases D will be to the right of C and in half of these cases to the left
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Re: Permutation Problem [#permalink]
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Bunuel wrote:
tania wrote:
In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D?

A.1680
B.2160
C.2520
D.3240
E.3360

Can someone explain how I should approach to solve the above problem?


We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\).

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\)

Answer: A.



Cannot be better than this one. +1.
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Re: Probability Q [#permalink]
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A

1) the total number of arrangements: 8!
2) excluding double counting (A1, A2 and A2, A1 are the same): 8!/2!*3! = 3360
3) the number of arrangements with C D is equal the number of arrangements with D C. Therefore, answer is 3360/2 = 1680.
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Re: Probability Q [#permalink]
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walker wrote:
A

1) the total number of arrangements: 8!
2) excluding double counting (A1, A2 and A2, A1 are the same): 8!/2!*3! = 3360
3) the number of arrangements with C D is equal the number of arrangements with D C. Therefore, answer is 3360/2 = 1680.


Can you be more clear in your explanation with the step 3?

I considered C to the right of D, the combination together as one unit and there are 7 units to be arranged with 2 A's and 3 B's. Obviously, I was not even close to any of the options. What's wrong with my approach?
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Re: Probability Q [#permalink]
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3) We always have twins, for example ???C?D?? - ???D?C??. In other words, exactly half of all arrangements has C to the right of the letter D.

Your approach considers only arrangements in which D and C are together. But we can have letters between D and C.
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Re: Probability Q [#permalink]
We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: .

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be

CAN ANYONE EXPLAIN LAST STEP

THANKS IN ADVANCE
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Re: Probability Q [#permalink]
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anilnandyala wrote:
We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: .

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be

CAN ANYONE EXPLAIN LAST STEP

THANKS IN ADVANCE


Obviously C and D can have ONLY TWO positions: C to the right of C OR to the left, how else?

Now, why should C (or D) be in more cases to the right (or to the left) of D (C)? Does probability favors either of these letters? No. Hence exactly in half of these cases D will be to the right of C and in half of these cases to the left.

Hope it's clear.
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Re: Probability Q [#permalink]
Bunuel.. I am confused here.

How do you say that in all the 3360 cases C and D will be together?
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Re: Probability Q [#permalink]
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prashantbacchewar wrote:
Bunuel.. I am confused here.

How do you say that in all the 3360 cases C and D will be together?


I'm not saying that anywhere.

"We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\).

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\)."
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Re: Probability Q [#permalink]
Thanks Bunuel Got the point
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Re: Probability Q [#permalink]
prashantbacchewar wrote:
Bunuel.. I am confused here.

How do you say that in all the 3360 cases C and D will be together?



In your approach you are thinking that D and C should be together But C can be anywhere after D.I mean If D is at the first position C can be at any position 2nd,3rd,4th or so on.

Consider KUDOS if u like the explanation.
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Re: Probability Q [#permalink]
Out of 8 letters, 2 A's and 3 B's can be arranged in 2! and 3! ways respectively.
Hence total ways = 8! / 2! * 3!
Now in half of them C will be towards the left of D. So for eliminating that we divide by 2.
8! / 2! * 3! *2 = 1680. Hence A.
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Re: Probability Q [#permalink]
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another way to think of it is you have 8 spaces, and you want to place C and D somewhere in those 8 spaces.

if you do 8c2 * 2! for example, you're saying there are 8 spots, i want to choose two of them for C and D. Also, i want to multiple by 2! because that is # of ways I can arrange C and D, ie C D, D C. But then you realize, you don't want to re-arrange C and D, you want C to the right of D, avoiding the re-arranging means just getting rid of the 2!, same as dividing by 2

(8c2 *2!) / 2

these are all the cases that C is to the right of D, and as Bunuel pointed out, it's also the same # of cases as D to the right of C.

You can just get the total cases now and subtract the above from it.
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Re: Permutation Problem [#permalink]
Bunuel wrote:
tania wrote:
In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D?
A.1680
B.2160
C.2520
D.3240
E.3360

Can someone explain how I should approach to solve the above problem?


We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\).

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\)

Answer: A.


Hi Bunnel,

If the question were:

Cases in which A is to the right of C?
tot=8!/2!*3!=3360

A to right of C = 3360/(2!)^2?
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Re: Permutation Problem [#permalink]
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cumulonimbus wrote:
Bunuel wrote:
tania wrote:
In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D?
A.1680
B.2160
C.2520
D.3240
E.3360

Can someone explain how I should approach to solve the above problem?


We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \(\frac{8!}{2!3!}=3360\).

Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \(\frac{3360}{2}=1680\)

Answer: A.


Hi Bunnel,

If the question were:

Cases in which A is to the right of C?
tot=8!/2!*3!=3360

A to right of C = 3360/(2!)^2?


Both A's or just one A? Anyway, in this case the problem will be out of the scope of the GMAT, so I wouldn't worry about it.

Questions about the same concept to practice:
susan-john-daisy-tim-matt-and-kim-need-to-be-seated-in-130743.html
meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html
goldenrod-and-no-hope-are-in-a-horse-race-with-6-contestants-82214.html

Hope it helps.
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