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CAT 2025 Lesson : Permutations & Combinations - Slotting Technique

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2.3 Slotting Techniques

In these questions, we would be given two or more groups, say for example Groups A and B have
aa and bb 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!a! ways. We will then have a+1a + 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 bb slots and then arrange them in b!b! ways.

Example 16

44 boys and 88 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 1212 be seated?
(
11) 8!×8P48! \times ^{8} \text{P}_{4}          (22) 8!×8C48! \times ^{8} \text{C}_{4}         (33) 8!×9P48! \times ^{9} \text{P}_{4}          (44) 8!×9C48! \times ^{9} \text{C}_{4}        

Solution

Permutation for seating the 88 girls = 8!\boldsymbol{8!} ways

We now have
99 slots between these 88 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^{9} \text{P}_{4} ways.

__ G __ G __ G __ G __ G __ G __ G __ G __

Total permutations =
8!×9P48! \times ^{9} \text {P}_{4}

Answer: (
33) 8!×9P48! \times ^{9} \text{P}_{4}


Example 17

In how many ways can 44 girls and 44 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!4! ways. We now have 4+14 + 1 = 55 slots to permute the 44 boys in.

Total permutations =
4!×5P44! \times ^{5} \text{P}_{4} = 24×12024 \times 120 = 2880\boldsymbol{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\boldsymbol{2} such arrangements.

We can seat the boys in the
44 positions in 4!\boldsymbol{4!} ways and the girls in their respective positions in 4!\boldsymbol{4!} ways.

Total permutations =
4!×4!×2 4! \times 4! \times 2 = 1152\boldsymbol{1152}

Answer: (I)
28802880; (II) 11521152


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