JEE Mathematics 2023 Question with Solution

Given: The word UNIVERSE.

Find: The number of $4$-letter words containing $2$ vowels and $2$ consonants without repetition.

The vowels are $I, U, E$ and the consonants are $N, V, R, S$.

Choose $2$ vowels from $3$ distinct vowels and $2$ consonants from $4$ consonants:

$${}^3C_2 \times {}^4C_2$$

Arrange the chosen $4$ letters in all possible ways:

$$4!$$

Therefore,

$${}^3C_2 \times {}^4C_2 \times 4! = 3 \times 6 \times 24 = 432$$

So, $432$ four letters word can be made.

Given: The letters of UNIVERSE with vowels $E, E, I, U$ and consonants $N, V, R, S$.

Find: The number of $4$-letter words with $2$ vowels and $2$ consonants without repetition.

Since repetition is not allowed, the repeated letter $E$ cannot be used twice as the same chosen position from identical letters. So we count using the distinct vowels $E, I, U$.

Select $2$ vowels:

$$\binom{3}{2} = 3$$

Select $2$ consonants:

$$\binom{4}{2} = 6$$

Arrange the selected $4$ letters:

$$4! = 24$$

Hence,

$$\binom{3}{2} \cdot \binom{4}{2} \cdot 4! = 3 \cdot 6 \cdot 24 = 432$$

Therefore, the required number of words is $432$.