Let the 6 people be represented as (A, a), (B, b), and (C, c)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
What if the question changed to all the male has to sit together?Brent@GMATPrepNow wrote:Let the 6 people be represented as (A, a), (B, b), and (C, c)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
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
Let's start by "gluing" the 3 men together. This ensures that they're seated together.ziyuenlau wrote: What if the question changed to all the male has to sit together?
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.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
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
Okay, let's imagine that the seats are arranged in order from left to right: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.
