JEE Mathematics 2024 Question with Solution

Given: The word DISTRIBUTION contains the letters: $I, I, I, T, T, D, S, R, B, U, O, N$.

Find: The total number of distinct $4$-letter words that can be formed.

Count arrangements by cases based on repetition.

1. Case 1: Three letters the same and one different, $(a,a,a,b)$

$$\binom{4}{1} \times \frac{4!}{3!} = 32$$

2. Case 2: Two letters each repeated twice, $(a,a,b,b)$

$$\frac{4!}{2! \cdot 2!} = 6$$

3. Case 3: Two identical letters and two distinct ones, $(a,b,c,c)$

$$\binom{2}{1} \times \binom{2}{1} \times \frac{4!}{2!} = 672$$

4. Case 4: All letters different, $(a,b,c,d)$

$$\binom{4}{1} \times 4! = 3024$$

Adding all cases,

$$\text{Total} = 3024 + 672 + 6 + 32 = 3734$$

Therefore, the total number of words is $3734$. Hence, the correct option is A.

Given: The letters in the word DISTRIBUTION are $I, I, I, T, T, D, S, R, B, U, O, N$.

Find: Number of distinct words of length $4$.

The repeated letters are $I$ occurring thrice and $T$ occurring twice. So possible repetition patterns for a $4$-letter word are taken casewise.

• Selection $(a,a,a,b)$:

$$\binom{4}{1} \times \frac{4!}{3!} = 32$$

• Selection $(a,a,b,b)$:

$$\frac{4!}{2! \cdot 2!} = 6$$

• Selection $(a,b,c,c)$:

$$\binom{2}{1} \times \binom{2}{1} \times \frac{4!}{2!} = 672$$

• Selection $(a,b,c,d)$:

$$\binom{4}{1} \times 4! = 3024$$

Now sum all these values:

$$\text{Total} = 3024 + 672 + 6 + 32 = 3734$$

So, the required number of words is $3734$.