How many different 3 letter words can be formed with the letters of word ‘JAIPUR’ when A and I are always to be excluded?

Answer

To solve the problem of how many different 3-letter words can be formed with the letters of the word "JAIPUR" when A and I are always excluded, we can follow these steps:

### Step 1: Identify the letters available
The letters in the word "JAIPUR" are J, A, I, P, U, R. Since A and I are to be excluded, we will only consider the letters J, P, U, and R.

**Available letters:** J, P, U, R

### Step 2: Count the remaining letters
After excluding A and I, we have the following letters left: J, P, U, R. This gives us a total of 4 letters.

**Remaining letters count:** 4 (J, P, U, R)

### Step 3: Choose 3 letters from the remaining 4
We need to select 3 letters from the 4 available letters (J, P, U, R). The number of ways to choose 3 letters from 4 can be calculated using the combination formula:

\[ \text{Number of ways to choose 3 letters from 4} = \binom{4}{3} \]

Using the combination formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\):

\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3! \cdot 1!} = \frac{4 \times 3!}{3! \cdot 1} = 4 \]

### Step 4: Arrange the chosen letters
Once we have chosen 3 letters, we can arrange them in different ways. The number of arrangements of 3 letters is given by \(3!\):

\[ 3! = 3 \times 2 \times 1 = 6 \]

### Step 5: Calculate the total number of different 3-letter words
Now, we multiply the number of ways to choose the letters by the number of ways to arrange them:

\[ \text{Total number of different 3-letter words} = \binom{4}{3} \times 3! = 4 \times 6 = 24 \]

### Final Answer
Thus, the total number of different 3-letter words that can be formed with the letters of the word "JAIPUR" when A and I are always excluded is **24**.

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