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Difficulty: MediumCounting with Restrictions and Repetitions

A bookshelf is to be arranged using 3 identical Mathematics books, 2 identical Physics books, and 1 Chemistry book. In how many distinct ways can all 6 books be arranged in a single row such that the Chemistry book is not adjacent to any Mathematics book?

Answer: 12

Answer

12
The single Chemistry book must not touch any of the 3 Mathematics books. This leaves three mutually exclusive possibilities for the Chemistry book: at the left end preceded by a Physics book (4 ways), at the right end followed by a Physics book (4 ways), or sandwiched between both Physics books as a single block (4 ways). Summing these gives 12 total valid arrangements.

Step-by-Step Solution

1
Analyze the placement restriction on the Chemistry book
The Chemistry book (C) cannot be placed next to any Mathematics book (M). Thus, C can only be adjacent to Physics books (P) or placed at the boundary of the row touching a P.
Eliminating adjacency to M restricts C to touch only P or the ends of the shelf.
2
Count valid arrangements when C is at the far left (position 1)
The row begins with C-P. The remaining 4 spots must be filled with 3 identical M's and 1 P, yielding 4! / (3! 1!) = 4 distinct ways.
Position 1 has only one neighbor (position 2), which must be P.
3
Count valid arrangements when C is at the far right (position 6)
The row ends with P-C. The remaining 4 spots must be filled with 3 identical M's and 1 P, yielding 4! / (3! 1!) = 4 distinct ways.
Position 6 has only one neighbor (position 5), which must be P.
4
Count valid arrangements when C is in an interior position (positions 2 through 5)
C must be sandwiched between two P's, forming the block (P-C-P). Arranging this single block along with the 3 identical M's (total of 4 items) yields 4! / (3! 1!) = 4 distinct ways.
Any interior placement requires both adjacent neighbors of C to be P.
5
Sum the counts from all mutually exclusive cases
Total valid arrangements = 4 + 4 + 4 = 12.
The three cases cover all possible non-overlapping valid placements for C.

Key Concept

Counting permutations with identical elements and positional restrictions using case analysis and block formation.
Estimated Time:1m 30s
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Topics

Counting with Restrictions and Repetitions

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