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Combinations Practice Question of the Week #27

For this week’s practice question, there’s only one correct answer. Good luck, we’ll be posting the answer and explanation tomorrow!

In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

  1. 24
  2. 48
  3. 56
  4. 72
  5. 96
Update: Here’s the answer/explanation post!

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10 Responses to Combinations Practice Question of the Week #27

  1. Robert December 1, 2011 at 6:03 am #

    Is this a correct approach?

    The total number of ways to organize A, B, C, D, and E is 5! = 120.

    This, however, has instances that break the rules by placing A+B together. If we reorganized these as A+B, C, D, and E, we can count all the rule-breaking instances as a 4! = 24.

    Subtract the rule-breakers from the total, and you get choice (E) 96.

    • Robert December 1, 2011 at 6:09 am #

      OK, I get it. All combinations of A+B, C, D, and E are rule-breakers. B+A, C, D, and E are also rule-breakers. 120-24-24 = 72, choice D.

  2. shri November 30, 2011 at 10:14 pm #

    96 seems to be the right answer but there has to be a better way to solve it than below:

    Total possibilities = 5! =120 ways
    if A,B are in positions 1&2 = 1x1x3x2= 6 ways
    similarly 2&3, 3&4, 4&5 = 6×3= 18 ways
    Total of 24 ways when A&B are together. Hence 120-24 =96 ways A&B are not together

    But I won’t get it right within the time frame on the test.

  3. chowdary November 30, 2011 at 8:57 pm #

    The answer is D(72)
    Without any restrictions in arrangement you get 5*4*3*2*1=120 configurations
    There are 2(4*3*2)=24 configurations in which A and B are next to each other
    Therefore the answer is 120-48=72

  4. MN November 29, 2011 at 3:57 pm #

    D. 72

  5. Dayam November 29, 2011 at 4:19 am #

    96. because there is one exception. Therefore, I multiplied 4X4X3X2X1=96.

    • Dennis November 29, 2011 at 3:47 pm #

      4x4x3x2 shouldnt work because there are two exceptions. If you picked either A or B for the first number [which is why you are going for number 4 instead of 5] then the next number cannot be the forbidden number [so you are picking from C,D,E]. Then you can pick any number from remaining.

      So 4 x 3 x 3 x 2 = 72

  6. Roshanak November 28, 2011 at 9:25 pm #

    i think the correct answer is E.96

  7. Dennis November 28, 2011 at 5:47 pm #


  8. lucym November 28, 2011 at 12:18 pm #

    answer is A-24

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