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Re: Permutation Problem [#permalink]
24 Dec 2009, 16:28

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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

Re: Permutation Problem [#permalink]
24 Dec 2009, 21:00

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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

Last edited by chetan2u on 25 Dec 2009, 00:10, edited 1 time in total.

Re: Permutation Problem [#permalink]
25 Dec 2009, 04:30

1

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Expert's post

gmatJP wrote:

why divide by only 2!3! ? there are 8spaces(letters) so dont you divide by 8!?

Permutations of n things of which P1 are alike of one kind, P2 are alike of second kind, P3 are alike of third kind ...................... Pr are alike of r th kind such that: P1+P2+P3+..+Pr=n is:

\frac{n!}{P1!*P2!*P3!*...*Pr!}.

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is 6!/2!2!, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!/4!3!2!.

In the original question there are 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.

LM wrote:

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

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.

Re: In how many different ways can the letters A, A, B [#permalink]
26 Jun 2013, 14:10

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Possible arrangements: 1) xcxxxxxx ---- 1C1 * 6!/(2!*3!) (D can be placed in 1 position) 2) xxcxxxxx ---- 2C1 * 6!/(2!*3!) (D can be placed in 2 positions) 3) xxxcxxxx ---- 3C1 * 6!/(2!*3!) (D can be placed in 3 positions) 4) xxxxcxxx ---- 4C1 * 6!/(2!*3!) (D can be placed in 4 positions) 5) xxxxxcxx ---- 5C1 * 6!/(2!*3!) (D can be placed in 5 positions) 6) xxxxxxcx ---- 6C1 * 6!/(2!*3!) (D can be placed in 6 positions) 7) xxxxxxxc ----- 7C1 * 6!/(2!*3!) (D can be placed in 7 positions)

Adding them all = (1+2+3+4+5+6+7) * 6!/(2!*3!) = 28 * 6!/(2!*3!) = 1680

Re: Permutation Problem [#permalink]
24 Dec 2009, 23:29

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

Re: Permutation Problem [#permalink]
25 Dec 2009, 19:30

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. _________________

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. _________________

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? _________________

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

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.

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." _________________

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.

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. _________________

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.

Re: Permutation Problem [#permalink]
26 May 2013, 17:46

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?

gmatclubot

Re: Permutation Problem
[#permalink]
26 May 2013, 17:46

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