Sunday, June 14, 2009

Problem 47 - GMAT Combinatorics

In how many ways can five girls stand in line if Maggie and Lisa cannot stand next to each other?

(A) 112
(B) 96
(C) 84
(D) 72
(E) 60


  1. 5! - (4)(2!)(3!)= 72, right?

  2. I agree that it's 72. A B C M L would be 5! without restrictions. Since M L can't be next to each other, you then need to subtract the 4*(3!) for whether (M L) is in the first, second, third or fourth position. In addition, multiply by 2! for L M versus M L.

  3. 5!-2x(4!)=72
    ML can be considered one block-hence number of arrangements = 2x4! (M and L can be L and M)
    Total number of arrangements 5!

  4. The answer is A.

    Total number of ways without restrictions is 5!=120

    Now just subtract the number of combinations where Lisa and Marie stand next to each other.
    1 2 3 4 5
    L M
    L M
    L M
    L M
    This is 4 combinations, now multiple by 4 by 2 since we can also do the opposite when Marie holds the first seat and Lisa the second.
    Total Number of combinations possible given restriction= 120-8=112 => A

  5. Anonymous, you are wrong, you forgot that movement of the other 3 people are also additional combinations, the answer is 72