JEE Mathematics 2023 Question with Solution

Given: The word is ASSASSINATION and all vowels must occur together.

Find: The number of words that can be formed using all letters such that the vowels occur together.

Vowels: $A, A, A, I, I, O$

Consonants: $S, S, S, S, N, N, T$

Treat all the vowels as one single group. Then the total entities become:

$$V, S, S, S, S, N, N, T$$

So there are $8$ entities in total.

The number of arrangements of these $8$ entities is

$$\frac{8!}{4!\,2!}$$

because there are $4$ identical $S$'s and $2$ identical $N$'s.

Now arrange the vowels within the vowel group. The vowels $A, A, A, I, I, O$ can be arranged in

$$\frac{6!}{3!\,2!}$$

ways, because there are $3$ identical $A$'s and $2$ identical $I$'s.

Therefore, the required number of words is

$$\frac{8!}{4!\,2!} \times \frac{6!}{3!\,2!}$$ $$= 840 \times 60$$ $$= 50400$$

Therefore, the number of required words is $50400$.

Given: All letters of ASSASSINATION are to be used and all vowels must be consecutive.

Find: The total number of distinct arrangements.

1. First separate vowels and consonants.

• Vowels: $A, A, A, I, I, O$
• Consonants: $S, S, S, S, N, N, T$

2. Group all vowels into one block. Then we arrange:

$$1 \text{ vowel block} + 7 \text{ consonants} = 8 \text{ entities}$$

3. Count arrangements of these $8$ entities. Since among consonants, $S$ repeats $4$ times and $N$ repeats $2$ times,

$$\text{Arrangements} = \frac{8!}{4!\,2!} = \frac{40320}{24 \times 2} = 840$$

4. Count internal arrangements of the vowel block. The vowels are $A, A, A, I, I, O$, so

$$\text{Internal arrangements} = \frac{6!}{3!\,2!} = \frac{720}{6 \times 2} = 60$$

5. Multiply the two counts.

$$\text{Total} = 840 \times 60 = 50400$$

Hence, the required numerical answer is $50400$.