The number of ways in which the letters AAABBCD can be arranged so that two Bs are together but no two As are together is:

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Understand the Problem

The question asks for the number of arrangements of the letters in 'AAABBCD' where two B's are together and no two A's are adjacent. This can be solved using combinatorial methods.

Answer

$24$
Answer for screen readers

The number of arrangements of the letters in 'AAABBCD' where two B's are together and no two A's are together is $24$.

Steps to Solve

  1. Treat the two B's as a single unit Combine the two B's into one unit, calling it "BB". The letters to arrange are now: A, A, A, C, D, BB. This gives us 6 units total.

  2. Calculate arrangements without restrictions Calculate the arrangements of the 6 units (A, A, A, C, D, BB). The formula for arrangements of indistinguishable objects is:

$$ \text{Arrangements} = \frac{n!}{n_1! \times n_2! \times \ldots} $$

Here, we have 3 A's, 1 C, 1 D, and 1 BB:

$$ \text{Arrangements} = \frac{6!}{3! \times 1! \times 1! \times 1!} $$

  1. Calculate the total arrangements Calculate the factorial values:

$$ 6! = 720, \quad 3! = 6 $$

Now substitute these into the formula:

$$ \text{Arrangements} = \frac{720}{6} = 120 $$

  1. Arrange the A's with restrictions Now, we need to arrange the A's so that no two A's are adjacent.

  2. Place non-A letters First, arrange the letters without A's: (C, D, BB) gives us:

$$ \text{Arrangements without A's} = 3! = 6 $$

  1. Identify slots for A's When the letters (C, D, BB) are arranged, they create gaps to place the A's. There are 4 gaps: before C, between C and D, between D and BB, and after BB.

  2. Place A's in slots We need to place 3 A's in these 4 slots where no two A's are together. The number of ways to choose 3 slots from the available 4 slots is given by:

$$ \binom{4}{3} = 4 $$

  1. Combine results Now, multiply the arrangements of non-A letters by the combinations of A's placement:

$$ \text{Total arrangements} = 6 \times 4 = 24 $$

The number of arrangements of the letters in 'AAABBCD' where two B's are together and no two A's are together is $24$.

More Information

In combinatorics, this problem utilizes the concept of arranging letters with restrictions, showcasing how to effectively count arrangements by treating groups of letters as single units and calculating based on permitted placements.

Tips

  • Failing to recognize when to treat letters as groups can lead to incorrect arrangements.
  • Not considering all possible slots for the letters creates unnecessary overlaps, leading to incorrect counts.

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