singhsa wrote:How many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?
A. 2
B. 3
C. 4
D. 6
E. 12

Came across this question and could easily solve it by taking students as A and B. But is there a formula used in this sort of question in case the nos are more complex? Thnx

Oh, BTW, OA - 6

We need to select 3 desks in a row, out of the four desks, so that there is only one empty desk between the students. This can be done in 2 ways only. Or, we select all four desks, so that there are at max 2 empty desks between the students. This can be done in just 1 way. The ways add up to 3, and the two fellows may again be arranged in 2! = 2 ways, within their desks. The many different ways would then sum up to [spoiler]3 × 2 = 6.

D[/spoiler]

No, there's no formula designed for tailored or fragmented distribution of elements.

sanju09 wrote:

singhsa wrote:How many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?
A. 2
B. 3
C. 4
D. 6
E. 12

Came across this question and could easily solve it by taking students as A and B. But is there a formula used in this sort of question in case the nos are more complex? Thnx

Oh, BTW, OA - 6

We need to select 3 desks in a row, out of the four desks, so that there is only one empty desk between the students. This can be done in 2 ways only. Or, we select all four desks, so that there are at max 2 empty desks between the students. This can be done in just 1 way. The ways add up to 3, and the two fellows may again be arranged in 2! = 2 ways, within their desks. The many different ways would then sum up to [spoiler]3 × 2 = 6.

D[/spoiler]

No, there's no formula designed for tailored or fragmented distribution of elements.

Hey, thnx. I get the concept now, but what if in the GMAT, the question asks us to arrange 3 students in a row of 10 desks in a similar manner as the question above? It'll be tedious solving the question this way.

singhsa wrote:

sanju09 wrote:

singhsa wrote:How many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?
A. 2
B. 3
C. 4
D. 6
E. 12

Came across this question and could easily solve it by taking students as A and B. But is there a formula used in this sort of question in case the nos are more complex? Thnx

Oh, BTW, OA - 6

We need to select 3 desks in a row, out of the four desks, so that there is only one empty desk between the students. This can be done in 2 ways only. Or, we select all four desks, so that there are at max 2 empty desks between the students. This can be done in just 1 way. The ways add up to 3, and the two fellows may again be arranged in 2! = 2 ways, within their desks. The many different ways would then sum up to [spoiler]3 × 2 = 6.

D[/spoiler]

No, there's no formula designed for tailored or fragmented distribution of elements.

Hey, thnx. I get the concept now, but what if in the GMAT, the question asks us to arrange 3 students in a row of 10 desks in a similar manner as the question above? It'll be tedious solving the question this way.

hope nice things of GMAT please