In these questions, we would be given two or more groups, say for example Groups A and B have a and b members respectively. The question might require us to find out the cases where members of Group B are not next to each other.
For this, we need to first seat (permute) the members of Group A which is a! ways. We will then have a+1 slots (between the members and the two ends) where we can slot the members of group B, so that they are never adjacent to each other. Here we can select the b slots and then arrange them in b! ways.
Example 16
4 boys and 8 girls are to sit in a row. None of the boys should have another boy seated next to him. In how many ways can the 12 be seated?
(1) 8!×8P4
(2) 8!×8C4
(3) 8!×9P4
(4) 8!×9C4
Solution
Permutation for seating the 8 girls = 8! ways
We now have 9 slots between these 8 students and at the ends as shown below (where G represents a girl seated and __ represents an empty slot). We can permute the boys in these slots in 9P4 ways.
__ G __ G __ G __ G __ G __ G __ G __ G __
Total permutations = 8!×9P4
Answer: (3) 8!×9P4
Example 17
In how many ways can 4 girls and 4 boys be seated such that
(I) none of the boys sit next to each other?
(II) none of the boys sit next to each other and none of the girls sit next to each other?
Solution
Case I: We first place the girls in 4! ways. We now have 4+1 = 5 slots to permute the 4 boys in.
Total permutations = 4!×5P4 = 24×120 = 2880
Case II: Since neither the boys nor the girls sit next to each other, they sit in alternate positions. So, the seating from left, either starts with a girl (GBGBGBGB) or starts with a boy (BGBGBGBG). We have 2 such arrangements.
We can seat the boys in the 4 positions in 4! ways and the girls in their respective positions in 4! ways.
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