The number of permutation of all the letters AAAABBBC in which the A's appear together in a block 4 letters or the B's appear together in block of 3 letters is

Options

A. 44
B. 50
C. 60
D. 89

Answer

To solve the problem of finding the number of permutations of the letters AAAABBBC where the A's appear together in a block of 4 letters or the B's appear together in a block of 3 letters, we can break it down into two cases.

### Step-by-Step Solution:

**Step 1: Case 1 - A's together in a block of 4 letters**

1. Treat the four A's as a single entity or block. Thus, we can represent the letters as:
\[ (AAAA), B, B, B, C \]
This gives us a total of 5 entities to arrange: (AAAA), B, B, B, C.

2. The number of permutations of these 5 entities is given by the formula for permutations of multiset:
\[ \text{Permutations} = \frac{n!}{p_1! \times p_2! \times \ldots} \]
where \( n \) is the total number of items, and \( p_1, p_2, \ldots \) are the counts of indistinguishable items.

3. Here, we have:
- Total entities = 5
- Indistinguishable B's = 3

Therefore, the number of permutations is:
\[ \text{Permutations} = \frac{5!}{3!} = \frac{120}{6} = 20 \]

**Step 2: Case 2 - B's together in a block of 3 letters**

1. Now, treat the three B's as a single entity or block. Thus, we can represent the letters as:
\[ A, A, A, (BBB), C \]
This gives us a total of 5 entities to arrange: A, A, A, (BBB), C.

2. Again, using the formula for permutations of multiset:
- Total entities = 5
- Indistinguishable A's = 3

The number of permutations is:
\[ \text{Permutations} = \frac{5!}{3!} = \frac{120}{6} = 20 \]

**Step 3: Total Permutations**

1. Since the two cases are mutually exclusive (A's together or B's together), we can add the permutations from both cases:
\[ \text{Total Permutations} = 20 + 20 = 40 \]

### Final Answer:
The total number of permutations of the letters AAAABBBC where the A's appear together in a block of 4 letters or the B's appear together in a block of 3 letters is **40**.

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