Hi VJesus12,

We're asked to arrange a group of 3 adults and 4 children in 7 seats with adults in the first, third, and seventh seats. We're asked for the number of arrangements possible. This question is a variation on a standard Permutation question. To solve it, we have to go from space to space and keep track of the 'options' available for each (noting that once we place a person, there is one fewer person available for the next equivalent spot).

For the 1st spot, there are 3 options
For the 2nd spot, there are 4 options
For the 3rd spot, there are 2 options
For the 4th spot, there are 3 options
For the 5th spot, there are 2 options
For the 6th spot, there are 1 options
For the 7th spot, there are 1 options

Total arrangements = (3)(4)(2)(3)(2)(1)(1) = 144 possible arrangments

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

VJesus12 wrote:How many different ways are there to arrange a group of 3 adults and 4 children in 7 seats if adults must have the first, third, and 7 seats?

A. 12
B. 144
C. 288
D. 1,400
E. 5,040

The OA is B .

I didn't know how to use the permutations here. I need your help experts.

The 3 adults can be arranged in 3! = 6 ways, and the 4 children can be arranged in 4! = 24 ways, so the total number of arrangements is 6 x 24 = 144.

Answer: B