mushyyy wrote:
Find the ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy J and a girl Y who can't be put in adjacent seats:
1)BGBGBGBG= 4*4*3*3*2*2*1*1 *2= 4!*4!*2 total ways
2)JY BGBGBG= 7* 3*3*2*2*1*1 = 3!*3!*7
now I guess I have to multiplied 2) by 4 because:
-JY BGBGBG
-YJ BGBGBG
-JY GBGBGB
-YJ GBGBGB
4!*4!*2 -(3!*3!*7)*4= result
do you think am I right?
First of all, please post answer choices for PS questions. Also do no reword or shorten the stem.Next, I wouldn't worry about this question at all as it's out of the scope of the GMAT. On the GMAT combination/probability questions are fairly straightforward and there is no need to spent too much time on this kind of problems.
Anyway:
There are 4 boys and 4 girls. We should sit them in a row so that boys and girls are seated alternatively and one particular boy (B') and one particular girl (G') are not in adjacent seats.
1. 4 boys and 4 girls can be seated alternately in 4!*4!*2 ways:BGBGBGBG - 4!*4! (4 boys can be arranged in their slots in 4! ways and similarly 4 girls can be arranged in their slots in 4! ways);
GBGBGBGB - 4!*4! (the same here, we just start a row with a girl this time).
2. Let's count total # of ways B' and G' are in adjacent seats:Consider them to be one unit
{B'G'}If we start a row with a boy:
*BG*BG*BG* --> we can place {B'G'} in one of the 4 empty slots (notice that we are not violating alternately seating restriction in this case) and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 4*3!*3!;
If we start a row with a girl:
G*BG*BG*B --> we can place {B'G'} in one of the 3 empty slots and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 3*3!*3!;
Total for the unit
{B'G'}: 4*3!*3!+3*3!*3!=7*3!*3!
The same # of ways if we consider unit to be
{G'B'};
Therefore total # of ways where B' and G' are in adjacent seats is 2*7*3!*3!.
3. Finally,
{total # of ways where B' and G' are not in adjacent seats and boys and girls are seated alternatively} = {total # of ways where boys and girls are seated alternatively} - {total # of ways where B' and G' are in adjacent seats} = 4!*4!*2 - 2*7*3!*3! = 648.
Hope it's clear.