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.
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