Combination

Question

In how many ways can three A's, two B's and three C's be arranged if no two C's are adjacent? A. 240 B. 360 C. 420 E. 1080 F. 2040

Vemuri · Answer

I believe this is a permutations question, not combinations because we are dealing with arrangements.

I am guessing the answer is between D & E (E & F as stated in the answer choices), but still don't have a way to determine it. Will keep trying.

dumb.doofus · Answer

I think the answer should be 200. Thanks Brent for pointing out Login to see the pics.. If someone has a quicker way to solve this.. that would be awesome.. please do share.. For now here is the solution: Let's divide this into two portions: A] When the 3 C's are placed together Let's denote the 3 C's as R.. So the elements are AAABBR Permutation = 6!/(3!2!) = 60 ------------------- (1) B] When 2 of the C's are placed together Let's again denote the 2 C's by R so now the elements are: AAABBCR Now again there are two cases: 1. When R is placed as the first element or the last element. In both cases permutation will be same. So permutation for R _ _ _ _ _ _ _ or _ _ _ _ _ _ R will be P = (5*5*4*3*2*1)/(3!2!) = 50 So total permutation for first and last position = 50 + 50 = 100 ----- (2) 2. When R is placed anywhere in the middle i.e. 2nd, 3rd, 4th, 5th or 6th position.. permutation will be same.. so P = (5*4*4*3*2*1)/(3!2!) = 40 total permutation = 5 * 40 = 200 --------------------- (3) Total permutation = 8!/(3!2!3!) - (1) + (2) + (3) = 560 - 60 + 100 + 200 = 200 Thanks Brent for pointing out. I had missed to subtract.

PAB2706 · Answer

hi i solved but got the answer 350... :roll: as always might have done some silly mistake

Brent@GMATPrepNow · Answer

dumb.doofus wrote:I think the answer should be 360. Login to see the pics.. If someone has a quicker way to solve this.. that would be awesome.. please do share.. For now here is the solution: Let's divide this into two portions: A] When the 3 C's are placed together Let's denote the 3 C's as R.. So the elements are AAABBR Permutation = 6!/(3!2!) = 60 ------------------- (1) B] When 2 of the C's are placed together Let's again denote the 2 C's by R so now the elements are: AAABBCR Now again there are two cases: 1. When R is placed as the first element or the last element. In both cases permutation will be same. So permutation for R _ _ _ _ _ _ _ or _ _ _ _ _ _ R will be P = (5*5*4*3*2*1)/(3!2!) = 50 So total permutation for first and last position = 50 + 50 = 100 ----- (2) 2. When R is placed anywhere in the middle i.e. 2nd, 3rd, 4th, 5th or 6th position.. permutation will be same.. so P = (5*4*4*3*2*1)/(3!2!) = 40 total permutation = 5 * 40 = 200 --------------------- (3) Total permutation = (1) + (2) + (3) = 60 + 100 + 200 = 360 This is a valid solution, except I believe that you have solved for the number of arrangements such that 2C's are adjacent. So, we need to subtract this value from the total number of arrangements without restrictions. The total number of arrangements without restrictions is 8!/3!3!2! = 560 So, our final answer is 560 - 360 = 200 I'll post a slightly faster approach shortly.

agoyal2 · Answer

PAB2706, I am also getting 350 :roll:

Total Arrangements = 8! / (3!*2!*3!) = 560

When 2 C's are considered 1 = 7! / (3!*2!*2!) = 210
(The above also includes when all 3 C's are together...)

So total cases = 560-210 = 350 ??

Brent@GMATPrepNow · Answer

PAB2706 · Answer

hey agoyal..see one mistake that we have done is we have not concidered the possibility when all 3 C's come together.. i guess 560- 210+ when all 3C's come together shud give the answer...

i dnt know why..just getting a mental block..cant figure...

Brent@GMATPrepNow · Answer

agoyal2 wrote:PAB2706, I am also getting 350 :roll:

Total Arrangements = 8! / (3!*2!*3!) = 560

When 2 C's are considered 1 = 7! / (3!*2!*2!) = 210
(The above also includes when all 3 C's are together...)

So total cases = 560-210 = 350 ??

From your calculation, I believe that you have joined two C's and are treating it as one unit. This approach is fine, however you can't treat this entity (CC) as being identical to the remaining C. Based on your calculation of 7!/3!2!2!, it looks like you have done this.

However, the CC and the C are not identical.
For example, the arrangement AACCBBAC is different from AACBBACC

Given this, we should treat CC and C as unique to get 7!/3!2! = 420
But this figure isn't correct either since CC and C aren't entirely unique since ACCCABBA is the same as ACCCABBA

So, from our 420 we need to subrtact the number of ways in which all 3 Cs are together (60 ways).

We now have 360 (420-60) ways in which we can have two Cs adjacent, and now we can subtract this from 560 as you stated earlier.

Brent Hanneson - Creator of GMATPrepNow.com

dtweah · Answer

Brent Hanneson wrote:

Beautiful Brent!!. I fumbled the answer choices some how and omitted the 200. I discovered error but didn't want to make the correction just to see who might still derive the correct solution from raw principles.

In case you are not familiar with the formulas Brent used u can used the following prinicples.

Since The restriction concerns the Cs, forget about them and arrange the A's and B's. How many ways can we arrange 3A's and 2'B. YOu have 5 spots. If you put the 3's in the first 3 spots, you just have 2 ways for the B's so the number of ways is 5C2. Or if you put the B's first you would have 3 ways for the A's or 5C3=5C2=10. So generally the number of ways to arrange n A's and m B's is (n+m)Cn=(n+m)Cm. So we have

A A A B B

Now we want to put the c's between these. That is the only way no two C's will be adjacent. There are actually 6 spots for any C. Count the left of first A, right of last B and all spots between. This is 3 Cs vying for 6 spots or 6C3.

10 x 20=200.

Yes Brent you are a Giant!!

PAB2706 · Answer

O thats right Brent..thanks ..that is the mistake I was doing.

After some thot i arrived at this point tat the total ways for C being together will be 420 as you have shown but after that i got completely blocked..cudnt think further....

thanks a lot

cheers!!

dumb.doofus · Answer

Brent Hanneson wrote: Very nice approach Brent.. Many thanks for sharing..

agoyal2 · Answer

Thanks Brent!!

sureshbala · Answer

First 3 A's and 2 B's can be arranged in 5!/(3!x2!) = 10 ways.

Now there are 6 places where 3 C's can be placed so that no two of them are together. Also, since all C's look same, arranging 3 C's in 6 places is nothing but choosing 3 places from 6 places which can be done in 6C3 = 20 ways.

Hence total number of ways = 10 x 20 = 200

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abhinav85 · Answer

Answer is 200. What is th OA dtweah???

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