In how many different ways can the letters A, A, B

Question

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

Bunuel · Accepted Answer

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.

Bunuel · Answer

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 .

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.

gmatJP · Answer

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

chetan2u · Answer

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

LM · Answer

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

GMAT TIGER · Answer

Cannot be better than this one. +1.

walker · Answer

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.

BarneyStinson · Answer

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?

walker · Answer

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.

anilnandyala · Answer

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

Bunuel · Answer

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.

pinchharmonic · Answer

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.

Bunuel · Answer

Questions about the same concept to practice:

https://gmatclub.com/forum/susan-john-d ... 30743.html
https://gmatclub.com/forum/meg-and-bob- ... 58095.html
https://gmatclub.com/forum/six-mobsters ... 26151.html
https://gmatclub.com/forum/mary-and-joe ... 26407.html
https://gmatclub.com/forum/goldenrod-an ... 82214.html

Hope it helps.

mps8085 · Answer

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

pushpitkc · Answer

These are the arrangements possible where letter C is to the right of D
D******C 
D*****C*
D****C**
D***C***
D**C****
D*C*****
DC******(7 combinations when D is the first alphabet)
*D*****C
*D****C* 
*D***C**
*D**C***
*D*C****
*DC*****(6 combinations when D is the second alphabet)
**D****C
**D***C*
**D**C**
**D*C***
**DC****(5 combinations when D is the third alphabet)
The total arrangements possible are  7+6+5+4+3+2+1 = 28

The total ways in which alphabets A,A,B,B,B,C can be arranged are \frac{6!}{2!*3!} = \frac{6*5*4*3*2}{2*3*2} = 6*5*2 = 60

Therefore, the total ways in which the alphabets can be arranged when C is to the right of D is 28*60 =  1680(Option A)

jmwon · Answer

After reviewing this problem and this similar one that Bunuel suggested (https://gmatclub.com/forum/susan-john-d ... 30743.html) I finally understand the hidden trap in the wording of this question.

The way this question is worded made me assume that C had to be right next to D, which made the problem much more confusing than it actually is. In fact the wording is just that C has to be TO THE RIGHT OF D, which means C and D could be respectively in any position in the row as long as C was right of D's position.

Therefore just need to calculate total # of ways A, A, B, B, B, C, D, E can be arranged, and then divide by 2 since the probability of C being to the right of D occurs half of the time compared to D being to the right of C.

\(\frac{8!}{(2! * 3!)} = \frac{3360}{2} = 1680\)

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In how many different ways can the letters A, A, B

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In how many different ways can the letters A, A, B

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