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yrozenblum
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A group of 10 friends consists of 5 married couples. How many seating Updated on: 29 Oct 2023, 12:32
Originally posted by yrozenblum on 29 Oct 2023, 11:03.
Last edited by gmatophobia on 29 Oct 2023, 12:32, edited 1 time in total.
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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 or her spouse?

A. 45
B. 120
C. 252
D. 3,840
E. 14,400
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D
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gmatophobia
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A group of 10 friends consists of 5 married couples. How many seating 29 Oct 2023, 11:48
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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
Show more

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
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A group of 10 friends consists of 5 married couples. How many seating 14 Apr 2024, 10:20
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gmathero911
Bunuel any similar GMAT Club problems to this one?
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­Check the links below:
https://gmatclub.com/forum/a-committee- ... 94068.html (GMAT Prep)
https://gmatclub.com/forum/m06-183707.html (GMAT Club Tests)
https://gmatclub.com/forum/if-there-are ... .html(GMAT Club Tests) 
https://gmatclub.com/forum/given-that-t ... 74964.html
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https://gmatclub.com/forum/at-a-dance-t ... 45695.html
https://gmatclub.com/forum/at-a-dance-t ... 45694.html
https://gmatclub.com/forum/if-a-committ ... 88772.html
https://gmatclub.com/forum/a-group-of-1 ... 19717.html
https://gmatclub.com/forum/in-how-many- ... 33266.html
https://gmatclub.com/forum/if-4-people- ... 99055.html
https://gmatclub.com/forum/given-that-t ... 58640.html
https://gmatclub.com/forum/a-committee- ... 30617.html

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A group of 10 friends consists of 5 married couples. How many seating 14 Apr 2024, 09:43
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Bunuel any similar GMAT Club problems to this one?
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A group of 10 friends consists of 5 married couples. How many seating 06 Jun 2024, 05:41
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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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A group of 10 friends consists of 5 married couples. How many seating 06 Sep 2024, 04:18
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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
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gmatophobia
Can you please explain why you multiplied by 2^5 and not 2! ?
Thankyou.
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A group of 10 friends consists of 5 married couples. How many seating 06 Sep 2024, 06:56
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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.
Show more
­
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.
Show more
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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A group of 10 friends consists of 5 married couples. How many seating 15 Sep 2026, 22:27
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Assume the couples as a block. so there are 5 blocks so we can arrange then in 5! ways. then, each couple can be arranged in 2! within each block. so for 5 blocks, it will be 2!^5
So final calculation = 5!*2!^5=120*32=3840. Opt D
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