Models

A) 48

Say S1 S2 E1 E2 N1 N2 are the models,

So S1S2 can be arranged in 2 ways, and same with others.So, # of ways of arranging = 2*2*2 =8

S1S2, S2S1
E1E2, E2E1
N1N2, N2N1

and the pair of participants can be arranged in 6 different ways

S1S2_E1E2_N1N2
S1S2_N1N2_E1E2
E1E2_N1N2_S1S2
E1E2_S1S2_N1N2
N1N2_S1S2_E1E2
N1N2_E1E2_S1S2

Total # of ways = 6*8=48

Models from same continent can be arranged among themselves in 2! ways each.

3 continents - (2!)^3 = 8

The 3 continents itself can be arranged in 3! ways - 6

Total - 6*8=48 A

GmatKiss wrote:There are six different models that are to appear in a fashion show. Two are from Europe, two are from South America, and two are from North America. If all the models from the same continent are to stand next to each other, how many ways can the fashion show organizer arrange the models?

A) 48
B) 64
C) 24
D) 8
E) 72

I received a PM asking me to comment.
Here's a step-by-step approach:

Number of continents that could occupy the first two positions = 3.
Number of ways to arrange the two models from this continent = 2.
Number of remaining continents that could occupy the middle two positions = 2.
Number of ways to arrange the two models from this continent = 2.
Number of remaining continents that could occupy the last two positions = 1.
Number of ways to arrange the two models from this continent = 2.
To combine the options above, we multiply:
3*2*2*2*1*2 = 48.

The correct answer is A.