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39%
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A class of 30 students take up a full row at a movie theater. If there are six couples among the students and if each couple wants to sit together, in how many different ways can the students sit?
A. \(24!*6\)
B. \(24!*12\)
C. \(24!*2^6\)
D. \(25!*6!\)
E. \(24!*6!*2^6\)
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A. \(24!*6\)
B. \(24!*12\)
C. \(24!*2^6\)
D. \(25!*6!\)
E. \(24!*6!*2^6\)
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Bunuel
Number the students from 1 through 30.
Glue couples together: {1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}, {11, 12}.
We'd be left with 30 - 12 = 18 single students and together with couples' units we'd get 6 + 18 = 24 units: {1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}, {11, 12}, {13}, {14}, {15}, ..., {30}.
These 24 units can be arranged in 24! ways. Each couple within their unit can be additionally arranged in 2! ways (for example, {1, 2} or {2, 1}).
Therefore the final answer is \(24!*(2*2*2*2*2*2)= 24!*2^6\).
Answer: C.
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General Discussion
Posts: 543
Bunuel
Suppose couples are single people. Then number of ways in which they will sit is 24!
Now each couple can sit in 2 way. There are six couples. 2^6
Total no of ways = 24!x2^6
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GMAT 1: 800 Q51 V51
Posts: 2,161
A class of 30 students take up a full row at a movie theater. If there are six couples among the students and if each couple wants to sit together, in how many different ways can the students sit?
30 students
6 couples
6 couples = 12 students
30 - 12 = 18 single students
First arrange the couples and single students.
Since each couple sits together, each couple becomes 1 element in the arrangement.
Since there are 6 couples and 18 single students, there are 24 elements to be arranged.
Ways to Arrange 24 Elements = 24!
Then, within each arrangement of 24 elements, there are 2 ways to arrange each of the 6 couples.
Ways to Arrange the Couples Within Each Arrangement of 24 Elements = \(2^6\)
Total Number of Arrangements = \(24!*2^6\)
A. \(24!*6\)
B. \(24!*12\)
C. \(24!*2^6\)
D. \(25!*6!\)
E. \(24!*6!*2^6\)
Correct answer: C
_________________
30 students
6 couples
6 couples = 12 students
30 - 12 = 18 single students
First arrange the couples and single students.
Since each couple sits together, each couple becomes 1 element in the arrangement.
Since there are 6 couples and 18 single students, there are 24 elements to be arranged.
Ways to Arrange 24 Elements = 24!
Then, within each arrangement of 24 elements, there are 2 ways to arrange each of the 6 couples.
Ways to Arrange the Couples Within Each Arrangement of 24 Elements = \(2^6\)
Total Number of Arrangements = \(24!*2^6\)
A. \(24!*6\)
B. \(24!*12\)
C. \(24!*2^6\)
D. \(25!*6!\)
E. \(24!*6!*2^6\)
Correct answer: C
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