5 boys and 4 girls sit in a straight line. Find the number of ways in which they can be seated if 2 girls are together and the other 2 girls are also together but seprate from the first 2 ?
5 boys and 4 girls sit in a straight line. Find the number of ways in which they can be seated if 2 girls are together and the other 2 girls are also together but seprate from the first 2 ?
Dear student,
I have provided two methods. Choose whichever you prefer.
Method I: Fix two ordered pairs of girls AB,CD. Then there are 7! ways to arrange the five boys and these two ordered pair. Of course we might have ABCD together, so to rule that out subtract 6! Similarly, we must rule out another factor of 6! to exclude CDAB. Thus for these two ordered pairs there are
7!−2×6!=3600
suitable arrangements. Now, sticking to these pairs but varying the order gets us
4×3600=14,400 And, finally, we can change the pairs. Instead of AB we could have had ACor AD so, finally,
3×14,400=43,200 Method II: (closer to what you were trying) Arrange the kids as
−AB−CD−
Where A,B,C,D denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are 15 satisfactory ways to arrange the boys:
{0,5,0},{1,4,0},{0,4,1},{2,3,0},{0,3,2},{1,3,1},{3,2,0},{0,2,3},{2,2,1},{1,2,2},{3,1,1},{1,1,3},{2,1,2},{4,1,0},{0,1,4}
(I'm writing them all out because I believe you only counted 10 of these). We must then pick some permutation of the girls to fill the slots labeled A,B,C,D and pick some permutation of the boys to populate the dashed regions thus
15×4!×5!=43,200
Dear student,
I have provided two methods. Choose whichever you prefer.
Method I: Fix two ordered pairs of girls AB,CD. Then there are 7! ways to arrange the five boys and these two ordered pair. Of course we might have ABCD together, so to rule that out subtract 6! Similarly, we must rule out another factor of 6! to exclude CDAB. Thus for these two ordered pairs there are
suitable arrangements.
Now, sticking to these pairs but varying the order gets us
4×3600=14,400And, finally, we can change the pairs. Instead of AB we could have had ACor AD so, finally,
3×14,400=43,200Method II: (closer to what you were trying) Arrange the kids as
−AB−CD−Where A,B,C,D denote the four girls (unspecified) and the boys go in the dashed areas. We know the middle dashed area must contain at least a single boy. There are 15 satisfactory ways to arrange the boys:
{0,5,0},{1,4,0},{0,4,1},{2,3,0},{0,3,2},{1,3,1},{3,2,0},{0,2,3},{2,2,1},{1,2,2},{3,1,1},{1,1,3},{2,1,2},{4,1,0},{0,1,4}
(I'm writing them all out because I believe you only counted 10 of these). We must then pick some permutation of the girls to fill the slots labeled A,B,C,D and pick some permutation of the boys to populate the dashed regions thus
15×4!×5!=43,200
