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Bunuel any similar GMAT Club problems to this one?
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10 * 1, * ,8 * 1, *, 6 *1, *, 4 *1, *, 2 * 1 = 3840

M1 * W1, *, M2 * W2, *, M3 * W3, *, M4 * W4, *, M5 * W5 = 3840

Is this reasoning correct?

With this method, I don't end up multiplying by 2! (to consider when W is chosen first). Hence, the question.
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gmatophobia
yrozenblum
A group of 10 friends consists of 5 married couples. How many seating arrangements are possible for these 10 friends to be seated in a row of 10 seats with 1 person per seat if each person is seated next to his o her spouse?

A. 45
B. 120
C. 252
D. 3,840
E. 14,400
Let's assume that the 5 couples are


  • \(M_1 \quad W_1\)
  • \(M_2 \quad W_2\)
  • \(M_3 \quad W_3\)
  • \(M_4 \quad W_4\)
  • \(M_5 \quad W_5\)

The 5 couples can be placed in 5! ways. Within each group, the couple can interchange their seats. Hence, each couple can be placed in 2 ways. As there are 5 couples, the members in the group can be arranged in \(2^5\) ways

Total = \(5! * 2^5 = 3840\) ways

Option D
­
gmatophobia
Can you please explain why you multiplied by 2^5 and not 2! ?
Thankyou.
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babyulikeit
gmatophobia
yrozenblum
A group of 10 friends consists of 5 married couples. How many seating arrangements are possible for these 10 friends to be seated in a row of 10 seats with 1 person per seat if each person is seated next to his o her spouse?

A. 45
B. 120
C. 252
D. 3,840
E. 14,400
Let's assume that the 5 couples are




  • \(M_1 \quad W_1\)
  • \(M_2 \quad W_2\)
  • \(M_3 \quad W_3\)
  • \(M_4 \quad W_4\)
  • \(M_5 \quad W_5\)

The 5 couples can be placed in 5! ways. Within each group, the couple can interchange their seats. Hence, each couple can be placed in 2 ways. As there are 5 couples, the members in the group can be arranged in \(2^5\) ways

Total = \(5! * 2^5 = 3840\) ways

Option D
­
gmatophobia
Can you please explain why you multiplied by 2^5 and not 2! ?
Thankyou.
­
We have 5 couples (M1, W1); (M2, W2); (M3, W3); (M4, W4); (M5, W5). These 5 units can be arranged in 5! ways. In each couple, M and W can be arranged in 2 ways. Since there are 5 couples, we get 2 * 2 * 2 * 2 * 2 = 2^5. Therefore, the final answer is 5! * 2^5 = 120 * 32 = 3,840.

Answer: D.­
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Yes2GMAT
10 * 1, * ,8 * 1, *, 6 *1, *, 4 *1, *, 2 * 1 = 3840

M1 * W1, *, M2 * W2, *, M3 * W3, *, M4 * W4, *, M5 * W5 = 3840

Is this reasoning correct?

With this method, I don't end up multiplying by 2! (to consider when W is chosen first). Hence, the question.
Yes, that approach works without multiplying by 2 because the possibility that W or M is chosen first is accounted for by the 10, 8, 6, 4, and 2.
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