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JEE Mathematics 2025 Question with Solution

The number of ways, in which the letters A, B, C, D, E can be placed in the 88 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:

  • A

    58805880

  • B

    960960

  • C

    840840

  • D

    57605760

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: 55 distinct letters A, B, C, D, E are to be placed in 88 boxes arranged in 33 rows with capacities 3,2,33, 2, 3 respectively. No row should remain empty and at most one letter can be placed in a box.

Find: The total number of valid arrangements.

Let the numbers of letters placed in the top, middle, and bottom rows be x1,x2,x3x_1, x_2, x_3 respectively. Then

x1+x2+x3=5x_1 + x_2 + x_3 = 5

with

x1≥1,x2≥1,x3≥1x_1 \ge 1, \quad x_2 \ge 1, \quad x_3 \ge 1

and row capacities

x1≤3,x2≤2,x3≤3.x_1 \le 3, \quad x_2 \le 2, \quad x_3 \le 3.

So the valid distributions are

(3,1,1), (2,1,2), (2,2,1), (1,1,3), (1,2,2).(3,1,1), \ (2,1,2), \ (2,2,1), \ (1,1,3), \ (1,2,2).

The distribution (1,3,1)(1,3,1) is invalid because the middle row has only 22 boxes.

For (3,1,1)(3,1,1):

(53)×3!×(21)×2×3\binom{5}{3} \times 3! \times \binom{2}{1} \times 2 \times 3 =10×6×2×2×3=720.= 10 \times 6 \times 2 \times 2 \times 3 = 720.

By symmetry, (1,1,3)(1,1,3) also gives 720720 ways.

For (2,1,2)(2,1,2):

(52)×3P2×(31)×2×3P2\binom{5}{2} \times {}^3P_2 \times \binom{3}{1} \times 2 \times {}^3P_2 =10×6×3×2×6=2160.= 10 \times 6 \times 3 \times 2 \times 6 = 2160.

For (1,2,2)(1,2,2):

(51)×3×(42)×2!×3P2\binom{5}{1} \times 3 \times \binom{4}{2} \times 2! \times {}^3P_2 =5×3×6×2×6=1080.= 5 \times 3 \times 6 \times 2 \times 6 = 1080.

For (2,2,1)(2,2,1):

(52)×3P2×(32)×2!×3\binom{5}{2} \times {}^3P_2 \times \binom{3}{2} \times 2! \times 3 =10×6×3×2×3=1080.= 10 \times 6 \times 3 \times 2 \times 3 = 1080.

Adding all valid cases,

720+720+2160+1080+1080=5760.720 + 720 + 2160 + 1080 + 1080 = 5760.

Therefore, the number of valid arrangements is 57605760, so the correct option is D.

Common mistakes

  • Including the distribution (1,3,1)(1,3,1) is incorrect because the middle row has only 22 boxes. Always check row capacities before counting arrangements.

  • Treating rows with 33 boxes and 22 boxes as identical leads to wrong permutation counts. Use 3P2{}^3P_2 for placing 22 letters in a 33-box row, but only 2!2! for a 22-box row.

  • Counting only selections of letters and forgetting their placements inside the chosen boxes undercounts the answer. After choosing letters, multiply by the number of ways to arrange them in the available boxes.

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