Beat The GMAT Challenge Question – September 20, 2010

by on September 20th, 2010

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Congrats to Soham for being selected as this week’s winner of the Beat The GMAT Challenge Question.  Soham will receive Premium Access to Beat The GMAT Practice Questions as a prize.  Above you will find the full video explanation to the question below.  Before watching the video above, you may want to first review this video about Calculating Combinations in Your Head; the lessons in this video are referenced in the video above:

Time for another Beat The GMAT Challenge Question. Of those who respond with the correct answer, one person will be randomly selected to receive free access to Beat The GMAT Practice Questions. Deadline to enter is 5PM PDT, Tuesday, September 21, 2010. Good luck!

How many unique 12-letter “words” can be created using 8 A’s and 4 B’s such that no 2 B’s are adjacent? (For example, the word “ABAABABAAABA” is permissible, but the word “AABBABABAAAA” is not permissible, since it has 2 adjacent B’s)

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  • Total number of arrangements without restrictions is 12!.
    Total number of arrangements that voilate restrictions 10!

    Therefore with restrictions is 12!-10! . The answer is 12!-10!

  • the answer should be 126.

    method: 8 A's can be arranged in just one way. Now we have 9 places where 4 B's can go. 4 places out of 9 can be chosen in 126 ways. Therefore, the answer is 126.

    • Don't you consider that all the Bs are identical ?

    • Hello Adeel !

      Let me explain you. we have 8 A's and 4 B's and we need to figure out how many ARRANGEMENTS are possible such that no two B's are together.

      First let us arrange 8 A's. As all are identical we have just one way to arrange them.

      Here my advice is to actually write 8 A's on your note pad. Now see my friend we have nine places where 4 B's can go. just count.

      these 4 places can be chosen in 126 ways. apply combination formula ie 9C4.

      Now as all 4 B's are identical we don't need to worry about arranging those 4 B's ( after placing them in four of the nine places.) So, here is your answer friend. i would have further counted for the permutations if the 4 B's were not identical. But as they are identical the answer is 126.

    • Great Explanation. It should be 126.

  • Total number of arrangements of 12 letters of 8 A's and 4 B's:

    Then we have to deduct the combinations where two or three or four B's are standing next to each other:
    12!/8!4! - 110 - 81 - 8 = 296

  • 12!/8!4! - 9 (4b's arranged in a row) - 10 (3b's arranged in a row) - 11*9 (11 ways 2'bs arranged * 9 ways of arranging the remaining 2b's for each of 11 ways) = 317

    495 - 19 - 99 = 317

    • can you explain 2Bs thing

  • 126 in my opinion. Here is yet another method:

    Y B X B X B X B Y , I need to fill in these X's and Y's with A's, Y's can have 0 number of A's also, so the sum of the number of As in these places will be 8.

    Let the number of A's in these places be x1, x2, x3, x4, x5 respectively.
    so we have,

    x1 + x2 + x3 + x4 + x5 = 8

    Now, x2, x3, x4 should be at least 1, otherwise the Bs would be together. So the new equation is

    x1 + (y2 + 1) + ( y3 + 1 ) + ( y4 + 1) + y5 = 8,

    Here I replaced the original variable x2 by y2 + 1, where y2>=0 .

    So, final equation is

    x1 + y2 + y3 + y4 + x5 = 5 , and the number of solutions of this equation given x1, y2, y3, y4, x5 > = 0 is

    9C4 = 126

    • Congrats soham, you are this week's winner! We'll follow up to get you free Premium access to Beat The GMAT Practice Questions.

    • Sohan, could you please explain the last step in your solution. I am not able to understand this step.

      From the following equations:
      x1 + y2 + y3 + y4 + x5 = 5 ,
      x1, y2, y3, y4, x5 > = 0

      How can we conclude that and the number of solutions of this
      equation given is 9C4 = 126 ?

  • we have to deduct the combination where two or three or four B's are standing next to each other and filling gaps between A's or B's is not right as that will be Derangement (12 places being fixed in number i.e. have to ignore extreme left/right places). Best way is to tie the redundant B's in a string, so BBBB as "X" or BB BB as "X X" and BBB B as "X B". In this way, we deal with straight formula arranging X-8A's, XX8A's,XB8A's..all of which are different. And lastly..we know in X, its only way arrangement.
    12!/8!4! - 9!/8! - 10!/8!2! - 9!/8!

  • Let the 8 As be written down in a line with one space between every adjacent A, and a space before the first A and after the last A. This would look as follows:

    _ A _ A _ A _ A _ A _ A _ A _ A _

    This arrangement of As could be made in only 1 possible way. Now, in order for none of the Bs to be adjacent to each other, we need to place 4 Bs in the spaces shown above. Because there are 9 spaces, and 4 Bs, we need to choose 4 out of these 9 spaces where we would like to place our Bs.

    This could be done in 9C4 = 9!/(4! X 5!) = 126 ways.

    Therefore, total no. of required combinations = No. of possible arrangements of As X No. of possible arrangements of Bs = 1 X 126 = 126 ways.

    Therefore, IMO, Answer is 126

  • Hi,

    The answer is 126.

    The simplest way to approach these questions is, first of all arrange those things on which there is no constraint in the problem.

    In this question, there is no constraint on the arrangement of a.

    Number of ways of arranging the 8 identical a's = 1

    After the arrangement of a's is over, it will look loke this


    So, we can clearly see that, 9 spaces are created in the arrangement of 8 a's. So to arrange 4 b's, we can select 4 spaces out of these 9 spaces.

    So, number of ways to arrange 4 identical b's in 9 places = 9C4*1

    So, total number of ways of arranging a's and b's together = (No of ways of arranging a) * (No of ways of arranging b)

    1*9C4 = 126

  • Hi,

    The answer is 126.

  • The simplest way to approach these questions is to arrange first those things on which no constraint is given in the problem.

    In this question, there is no constraint on the arrangement of a's, so arrange a's first.

    No of ways of arranging the 8 identical a's = 1

    Now, after the arrangement of a's, 9 spaces are created among them:


    As we can clearly see that 9 spaces are created after the arrangement of 8 a's. Now arrange 4 b's in these 9 spaces so that the constraint of no 2 b's together is not violated:

    No of ways of arranging 4 identical b's in 9 spaces = 9C4*1

    Total number of ways of arranging all a's and b's = (No of ways of arranging the 8 identical a's) * (No of ways of arranging 4 identical b's) = 1*9C4 = 126

  • Thyank you S.Sharma i got it. The explanation was good and it was further helped by Sunil.

  • all of u took gr8 care in not taking two Bs together but no body paid heed to the arrangements wen two or more A can come together
    so it shud go the following way
    when no As ttogether like XAXAXAXAXAXAXAXAX
    9 places for B = 9C4=126
    when 2 As together consider two A as one entity like XAXAXAXAXAXAXAAX we have only 8 places for B now =8C4= 70
    similarly when 3As together then 7C4= 35
    when 4 As together then 6C4= 15
    when 5 as together then 5C4 = 5
    and lastly when 6 as together then 4C4=1
    summing up all gives 252

    • Hi Neha,

      While considering the case in which you have calculated 9C4 = 126, these cases takes care of all the conditions when 2 or 3 or 4 or 5 or 6 a's are together.

      Because, the spaces created among them are 9, and you are filling only 4 spaces. Consider the below example:


      In this case 4 spaces have been used by b, but in the remaining spaces which are not filled by b will be occupied by a.

      So 126 is the correct answer..

  • The answer should be 126 because as 8A will generate 9 places to move so 4Bs will take place in any 4 of these 9 spaces. So the answer is to be 9C4=126.
    Moreover there is one alternative solution that is, the arrangements where Bs are adjacent could be BB-BB, BBB-B, BBBB. If we start with BB-BB and take each BB as a group then we'd have 8As and 2BBs which would make 10C2=45. Likewise BBB-B arragement can occur with 8As and BBB-B form. This would make 10p2=90 but in this arrangement if BBB touches B then we'd have 4Bs and BBBB arrangement would be double counted. As we have 9 places, we have 9X2=18 extra arragements which will be subtracted from 90 makes 72. Finally BB-B-B arragement 9 places X (9-1)C2=252 because 8 spaces will be left for B-B to move. So adding 90+72+252=369 arrangements B is adjacent to other B. The number of all arrangements 12C8=495 and 495-369=126 will be the number of all arrangements in which no B is adjacent to other B.

    • sorry 90+72+252 will be 45+72+252

    • i got it..thanks sunil it must be 126

  • is anybody interested in making study group specially for 3 months prep

    • Neha, where r u located? We do a meetup group in Virginia...lemme know if you are interested...

  • 8 As have 9 spaces between them (including the first spot)


    so we have 9 spots and 4 Bs ... the answer is 9c4.

    Cheers !!

  • 9 empty spots to be filled by 4 Bs.
    Answer: 9C4

  • Easy one. 8 A's and there are 9 places between them to fill the B's without two B's being next to each other.

    9 places to fill in 4 B's -- 9C4 -- 126 ways.

  • The answer is 126 ways.


    You have 9 possible places to put the 4 B´s, therefore 9C4 = 126

  • 126

  • 9c4 = 126

  • if it was 7 instead of 12 the answer is 1(!)
    if it was 8 the answer is 1(!)+2(!)
    therefor for 12 the answer is 1(!)+[1(!)+2(!)]+[1(!)+2(!)+3(!)]...

    which equals 1+4+10+20+35+56 = 126

  • Still not getting, why in 8As.. two A's can not come together. viz in lieu of _A_A_A_A_A_A_A_A_ .... I still can have AAA_AA_A_A_A. Basically my doubt was left-right of a B is there, For A its not.

  • Can someone explain me that _A_A_A_A_A_A_A_A_ vs AAA_A_AA_A_A. In all above way.. it seems there is only one A left-right of B, my doubt was we can have multiple A's either side of a lone B.

    • Hi Lalit,

      In _A_A_A_A_A_A_A_A_ , we are arranging the 4 b's in these 9 spaces. So the remaining 5 spaces in which b is not arranged will be eliminated and a's will be together for those cases.

      For example:

      Counting the spaces from left, lets assume for one particular case b's took the 1st,3rd,5th and 7th space. So out resulting case will be like :


      So the remaining spaces in which b's are not place, a's are put together in those spaces thereby eliminating those spaces as We are required to arrange only a's and b's .

  • My opinion
    The 'A' s will just bulk the 'B' in the spare places so we whould look just at the Bs. Let's count the different numbers:

    Total number:
    12*11*10*9 possible arrangement for Bs (= different numbers)

    Adjacent Bs:
    One group of 2 adjacents Bs is enough. This group has 11 possible places in a 12 digit number
    *10 (spaces left where to put a B)
    *9 for the last B

    That gives
    12*11*10*9 - 11*10*9
    =110*100 -110

  • I was wrong, the result is in fact bigger than 2^12 :P
    the Bs are the same, so
    all possible cases - BB cases( with BBB and BBBB duplicates and triplicates removed )
    C{12,4} - [11*C{10,2} - (10 * C{9,1} - 9)]
    495 - (396-90+9)

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