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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?
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.
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. 4*2=8 Total Number of combinations possible given restriction= 120-8=112 => A
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5! - (4)(2!)(3!)= 72, right?
ReplyDeleteI 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.
ReplyDelete5! - 4*2!*31 = 72
ReplyDelete5!-2x(4!)=72
ReplyDeleteML can be considered one block-hence number of arrangements = 2x4! (M and L can be L and M)
Total number of arrangements 5!
The answer is A.
ReplyDeleteTotal 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.
4*2=8
Total Number of combinations possible given restriction= 120-8=112 => A
Anonymous, you are wrong, you forgot that movement of the other 3 people are also additional combinations, the answer is 72
ReplyDelete