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Three couples need to arranged in a row for a group photo. I

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by hazelnut01 » Sat May 20, 2017 6:54 am
Three couples need to arranged in a row for a group photo. If the couples cannot be separated, how many different arrangements are possible?

A. 6
B. 12
C. 24
D. 48
E. 96
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by Brent@GMATPrepNow » Sat May 20, 2017 7:02 am
ziyuenlau wrote:Three couples need to arranged in a row for a group photo. If the couples cannot be separated, how many different arrangements are possible?

A. 6
B. 12
C. 24
D. 48
E. 96
Let the 6 people be represented as (A, a), (B, b), and (C, c)

Take the task of arranging the 3 couples and break it into stages.

Stage 1: Choose the order of the A/a couple.
They can be arranged as either Aa or aA
So, we can complete stage 1 in 2 ways

Stage 2: Choose the order of the B/b couple.
They can be arranged as either Bb or bB
So, we can complete stage 2 in 2 ways

Stage 3: Choose the order of the C/c couple.
They can be arranged as either Cc or cC
So, we can complete stage 3 in 2 ways

Stage 4: Arrange the 3 couples
We can arrange n objects in n! ways.
So, we can arrange 3 couples in 3! ways (6 way)
We can complete this stage in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus arrange the 6 people) in (2)(2)(2)(6) ways ([spoiler]= 48 ways[/spoiler])

Answer: D
--------------------------

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by hazelnut01 » Sat May 20, 2017 7:10 am
Brent@GMATPrepNow wrote:
ziyuenlau wrote:Three couples need to arranged in a row for a group photo. If the couples cannot be separated, how many different arrangements are possible?

A. 6
B. 12
C. 24
D. 48
E. 96
Let the 6 people be represented as (A, a), (B, b), and (C, c)

Take the task of arranging the 3 couples and break it into stages.

Stage 1: Choose the order of the A/a couple.
They can be arranged as either Aa or aA
So, we can complete stage 1 in 2 ways

Stage 2: Choose the order of the B/b couple.
They can be arranged as either Bb or bB
So, we can complete stage 2 in 2 ways

Stage 3: Choose the order of the C/c couple.
They can be arranged as either Cc or cC
So, we can complete stage 3 in 2 ways

Stage 4: Arrange the 3 couples
We can arrange n objects in n! ways.
So, we can arrange 3 couples in 3! ways (6 way)
We can complete this stage in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus arrange the 6 people) in (2)(2)(2)(6) ways ([spoiler]= 48 ways[/spoiler])

Answer: D

Cheers,
Brent
What if the question changed to all the male has to sit together?
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by [email protected] » Sat May 20, 2017 10:20 am
Hi ziyuenlau,

In your last post, I assume you mean that the 3 men would have to sit side-by-side and the three women would sit in the remaining 3 seats. To answer that question, we have to think about the various possibilities (although you'll notice that the math is essentially the same with each calculation):

Seats 1-3 are men:
M M M F F F
(3)(2)(1)(3)(2)(1) = 36 arrangements

Seats 2-4 are men:
F M M M F F
(3)(3)(2)(1)(2)(1) = 36 arrangements

Seats 3-5 are men:
F F M M M F
(3)(2)(3)(2)(1)(1) = 36 arrangements

Seats 4-6 are men:
F F F M M M
(3)(2)(1)(3)(2)(1) = 36 arrangements

In this situation, there would be... 36+36+36+36 = 144 arrangements.

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by elias.latour.apex » Sat May 20, 2017 3:41 pm
The previous post is correct but overly complicated.

If we treat the group of men as a unit, we can ask ourselves how many different places the first member of that group could sit in. He could sit in seat 1, 2, 3, or 4. He could not sit in seat 5 or 6 because the group of 3 would surpass the bounds of the range.

So we start with 4 different possibilities. The women could sit in the 3 empty seats in 3! ways and the men could be arranged inside their group in 3! ways.

So the answer will be 4 x 3! x 3! = 144
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by Brent@GMATPrepNow » Sun May 21, 2017 4:30 am
ziyuenlau wrote: What if the question changed to all the male has to sit together?
Let's start by "gluing" the 3 men together. This ensures that they're seated together.
We now have 4 things to arrange: womanA, womanB, womanC, and the glued-together men
We can arrange these 4 things in 4! ways (= 24 ways)

For each of these 24 different arrangements, we can take the glued-together men and arrange the men in 3! ways.

So, the TOTAL number of arrangements = (24)(3!) = (24)(6) = 144

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by Jeff@TargetTestPrep » Fri May 26, 2017 10:39 am
hazelnut01 wrote:Three couples need to arranged in a row for a group photo. If the couples cannot be separated, how many different arrangements are possible?

A. 6
B. 12
C. 24
D. 48
E. 96
We can label the couples as [A-B], [C-D], and [E-F]. Since the couples cannot be separated, we see that there are "3 slots," and thus, the couples can initially be arranged in 3! = 6 ways.

However, since each couple cannot be separated, each couple can be individually arranged in 2! = 2 ways. Since there are 3 couples, those couples can be arranged in 2 x 2 x 2 = 8 ways.

Thus, the total number of ways to arrange the group is 6 x 8 = 48 ways.

Answers: D

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by hoppycat » Wed May 31, 2017 4:30 am
elias.latour.apex wrote:The previous post is correct but overly complicated.

If we treat the group of men as a unit, we can ask ourselves how many different places the first member of that group could sit in. He could sit in seat 1, 2, 3, or 4. He could not sit in seat 5 or 6 because the group of 3 would surpass the bounds of the range.

So we start with 4 different possibilities. The women could sit in the 3 empty seats in 3! ways and the men could be arranged inside their group in 3! ways.

So the answer will be 4 x 3! x 3! = 144

Can you explain hte part in red. I don't get the part about surpassing the bounds of the range.
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by elias.latour.apex » Wed May 31, 2017 3:45 pm
hoppycat wrote:
elias.latour.apex wrote:The previous post is correct but overly complicated.

If we treat the group of men as a unit, we can ask ourselves how many different places the first member of that group could sit in. He could sit in seat 1, 2, 3, or 4. He could not sit in seat 5 or 6 because the group of 3 would surpass the bounds of the range.

So we start with 4 different possibilities. The women could sit in the 3 empty seats in 3! ways and the men could be arranged inside their group in 3! ways.

So the answer will be 4 x 3! x 3! = 144

Can you explain hte part in red. I don't get the part about surpassing the bounds of the range.
Okay, let's imagine that the seats are arranged in order from left to right:

1 2 3 4 5 6

Where can the first man sit? If he sits in the first seat, it will look like this:

M M M W W W

He could also sit in the second seat.

W M M M W W

Or in the third or fourth seat.

W W M M M W || W W W M M M

But the man cannot sit in the 5th seat because there will not be enough room for the other two men to sit. So there are only four ways the group of men can sit.
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by hoppycat » Sun Jun 04, 2017 7:36 am
Ahhhhh thanks
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