Queue combinatorics

Sri,

The easiest way to approach this problem is to find the total number of ways 6 students can be arranged in a line and then subtract the case where two boys will be standing together.

The total number of ways six students can be arranged in a line = 6!

If two boys (Paul and Stuart) are always standing together (assume they are tied to each other) then total number of ways students can be arranged = 5! x 2 (multiplying by two because Paul standing in front of Stuart is separate from Stuart standing in front of Paul)

So case where they won't be standing next to each other = 6! - 5! x 2
= 6 x 5! - 5! x 2
= 5! (6 - 2)
= 5!4 Answer

Hope it helps..

gmattesttaker2 wrote:Hello,

Can you please assist with this:

Four girls and two boys will stand in line, in how many ways can they stand if the two boys must not stand next to each other?

A) (5!)(4)
B) (5!)(3)
C) (5!)(2)
D) (4!)(3)
E) (4!)(2)

OA: A

Alternate approach:

Number of ways to arrange the 6 children = 6!.

Number of pairs that can be formed from the 6 positions = 6C2 = (6*5)/(2*1) = 15.
Of these 15 pairs, 5 are composed of adjacent positions:
1,2
2,3
3,4
4,5
5,6
Thus, of the 15 pairs that could be occupied by the 2 boys, 10 are composed of NON-adacent positions.

Implication:
Since 10/15 = 2/3, the boys will occupy non-adjacent positions in 2/3 of the 6! possible arrangements:
(2/3) * (6*5*4*3*2*1) = 5! * 4.

The correct answer is A.

alt approach

-g-g-g-g- out of 5 "-" two boys can come at any place and remaining 4 girls can be arranged in 4! ways 4!x5x4 ----> 5!x4

6! - 2x5!

5!(6-2)

4x5!

{A}