TEZPUR 2025 — Mathematics PYQ

To solve this, let us analyze the word GRADER.

• Total letters: 6 (G, R, A, D, E, R)

• Vowels: A, E (2 vowels)

• Consonants: G, R, D, R (4 consonants)

• Positions: 1, 2, 3, 4, 5, 6 (Odd places: 1, 3, 5; Even places: 2, 4, 6)

The condition is that no vowel can occupy an odd place. Therefore, the 2 vowels must occupy the even places (2, 4, 6).

1. Selecting even places for vowels: There are 3 even places, and we have 2 vowels to place. The number of ways to choose and arrange the 2 vowels in these 3 places is $P(3,2)$. Since there are two 'R's in the consonants, we must account for repetitions.

   • Arranging vowels (A, E) in 3 even spots: $3\times 2=6$ ways.

2. Arranging consonants (G, R, D, R) in remaining 4 spots: There are 4 spots left. The letters are {G, R, D, R}. The number of arrangements is:

   $$\frac{4!}{2!}=\frac{24}{2}=12\text{ ways}$$

Total arrangements:

$$Total=6\times 12=72$$

Correction: Wait, re-evaluating the vowel placement. If we place 2 vowels in 3 even positions, the number of ways is ${}^3{P}_2=6$. For the consonants, we have 4 consonants including two R's, which is $4!/2!=12$. The total is $6\times 12=72$.

Looking at the options provided, if the calculation results in 72 and it is not listed, let's re-verify the vowel/consonant count. Word: G-R-A-D-E-R. Vowels: A, E (2). Consonants: G, R, D, R (4). If the vowels must be in even places (2, 4, 6), then the calculation holds. Given the standard options in such problems, the closest logical approach leads to (b) 48 if the constraint was interpreted differently (e.g., only 2 even places available).

Assuming the standard combinatorics approach:

$$Ways=P(3,2)\times \frac{4!}{2!}=6\times 12=72$$