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Difficulty: MediumCounting with Restrictions and Repetitions

A security analyst needs to generate a 6-symbol access code using all of the following characters: A,A,A,B,B,A, A, A, B, B, and CC. How many distinct 6-symbol access codes can be formed such that the two BB's are not adjacent to each other?

  1. A
    20
  2. B
    24
  3. 40Answer
  4. D
    60
  5. E
    120

Answer

40 distinct 6-symbol access codes can be formed.
First arrange the four symbols that have no adjacency restrictions (A,A,A,A, A, A, and CC). The number of unique ways to arrange these 4 symbols with 3 identical AA's is 4!3!=4\frac{4!}{3!} = 4. Placing these 4 symbols in a row creates 5 available spaces (one before the first character, three between characters, and one after the last character). To ensure the two BB's are not adjacent, choose 2 distinct spaces out of the 5 for the BB's. The number of ways to select 2 spaces out of 5 is (52)=10\binom{5}{2} = 10. Multiplying the arrangement of base characters by the space selections yields 4×10=404 \times 10 = 40 total valid codes.

Step-by-Step Solution

1
Arrange the non-restricted characters
There are 4 ways to arrange the remaining characters A,A,A,A, A, A, and CC.
The number of distinct arrangements of 4 characters where 3 are identical is 4!3!×1!=4\frac{4!}{3! \times 1!} = 4.
2
Determine the available slots for the restricted character
Arranging 4 characters creates 5 distinct slots (including the two ends): _ X _ X _ X _ X _.
To ensure no two BB's are adjacent, each BB must occupy a distinct slot around or between the arranged characters.
3
Calculate the combinations of slot positions
There are (52)=10\binom{5}{2} = 10 ways to choose 2 slots for the two identical BB's.
Since the two BB's are identical, order of selection within the slots does not matter.
4
Apply the Fundamental Counting Principle
4×10=404 \times 10 = 40 distinct codes.
Multiply the number of ways to arrange the base characters by the number of ways to place the BB's into the available slots.

Key Concept

Counting with restrictions using the slotting method for non-adjacent elements
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Topics

Counting with Restrictions and Repetitions

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