The number of ways in which the letters AAABBCD be arranged so that two B's are together but no two A's are together is?

Steps to Solve

1. Group the Two B's Together

That means we can treat "BB" as a single unit. Thus, the letters we have to arrange are ( AAABCD ) which becomes ( AAABB(CD) ).

2. Count the Total Elements to Arrange

Now, we have the groups: ( X = BB, A, A, A, C, D ) (where ( X ) represents the grouped B's). Therefore, we will arrange ( X, A, A, A, C, D ).

This gives us 6 elements: ( 1BB, 3A, 1C, 1D ).

3. Calculate Arrangements with No Two A's Together

To ensure no two A's are together, we first arrange the other characters ( (BB, C, D) ). The arrangement of ( BB, C, D ):

The characters are 3 unique positions, so they can be arranged in ( 3! = 6 ) ways.

4. Insert A's in Between Other Characters

Once arranged, we create "gaps" around them:

• BB
• C
• D

This creates 4 possible gaps to place the A's: before BB, between BB and C, between C and D, and after D.

5. Choose Gaps for A's

We need to select 3 out of these 4 gaps to place the A's. The number of ways we can choose 3 gaps from these 4 gaps is given by:

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

6. Final Count of Arrangements

Multiply the arrangements of BB, C, D by the ways to place A's:

$$ Total , arrangements = (3!) \cdot \binom{4}{3} = 6 \cdot 4 = 24 $$