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NVAMediumJEE 2024Applications of P&C

JEE Mathematics 2024 Question with Solution

In an examination of Mathematics paper, there are 2020 questions of equal marks, and the question paper is divided into three sections: A, B, and C. A student is required to attempt a total of 1515 questions, taking at least 44 questions from each section. If section A has 88 questions, section B has 66 questions, and section C has 66 questions, then the total number of ways a student can select 1515 questions is:

Answer

Correct answer:11376

Step-by-step solution

Standard Method

Given: Section A has 88 questions, section B has 66 questions, and section C has 66 questions. A student must select 1515 questions with at least 44 from each section.

Find: The total number of valid selections.

Let xA,xB,xCx_A, x_B, x_C be the numbers of questions selected from sections A, B, and C respectively. Then

xA+xB+xC=15x_A + x_B + x_C = 15

with constraints

xA≥4,xB≥4,xC≥4x_A \ge 4, \quad x_B \ge 4, \quad x_C \ge 4

and also

xA≤8,xB≤6,xC≤6x_A \le 8, \quad x_B \le 6, \quad x_C \le 6

Now list all possible integer triples satisfying these conditions:

  • (7,4,4)(7,4,4)
  • (6,4,5)(6,4,5)
  • (5,4,6)(5,4,6)
  • (6,5,4)(6,5,4)
  • (5,5,5)(5,5,5)
  • (4,5,6)(4,5,6)
  • (5,6,4)(5,6,4)
  • (4,6,5)(4,6,5)

For each case, the number of selections is

(8xA)(6xB)(6xC)\binom{8}{x_A}\binom{6}{x_B}\binom{6}{x_C}

So,

(87)(64)(64)=8×15×15=1800\binom{8}{7}\binom{6}{4}\binom{6}{4} = 8 \times 15 \times 15 = 1800 (86)(64)(65)=28×15×6=2520\binom{8}{6}\binom{6}{4}\binom{6}{5} = 28 \times 15 \times 6 = 2520 (85)(64)(66)=56×15×1=840\binom{8}{5}\binom{6}{4}\binom{6}{6} = 56 \times 15 \times 1 = 840 (86)(65)(64)=28×6×15=2520\binom{8}{6}\binom{6}{5}\binom{6}{4} = 28 \times 6 \times 15 = 2520 (85)(65)(65)=56×6×6=2016\binom{8}{5}\binom{6}{5}\binom{6}{5} = 56 \times 6 \times 6 = 2016 (84)(65)(66)=70×6×1=420\binom{8}{4}\binom{6}{5}\binom{6}{6} = 70 \times 6 \times 1 = 420 (85)(66)(64)=56×1×15=840\binom{8}{5}\binom{6}{6}\binom{6}{4} = 56 \times 1 \times 15 = 840 (84)(66)(65)=70×1×6=420\binom{8}{4}\binom{6}{6}\binom{6}{5} = 70 \times 1 \times 6 = 420

Adding all these values,

1800+2520+840+2520+2016+420+840+420=113761800 + 2520 + 840 + 2520 + 2016 + 420 + 840 + 420 = 11376

Therefore, the total number of ways is 1137611376.

Using shifted variables

Given: xA+xB+xC=15x_A + x_B + x_C = 15 with xA,xB,xC≥4x_A, x_B, x_C \ge 4.

Find: All admissible distributions and the total number of selections.

Introduce

yA=xA−4,yB=xB−4,yC=xC−4y_A = x_A - 4, \quad y_B = x_B - 4, \quad y_C = x_C - 4

Then

yA+yB+yC=3y_A + y_B + y_C = 3

with yA,yB,yC≥0y_A, y_B, y_C \ge 0.

The non-negative integer solutions of

yA+yB+yC=3y_A + y_B + y_C = 3

are counted by

(3+3−13−1)=(52)=10\binom{3+3-1}{3-1} = \binom{5}{2} = 10

but not all of these are valid because section B and section C each have at most 66 questions, so xB,xC≤6x_B, x_C \le 6. After checking the upper bounds, the valid cases reduce to:

(7,4,4),(6,4,5),(5,4,6),(6,5,4),(5,5,5),(4,5,6),(5,6,4),(4,6,5)(7,4,4), (6,4,5), (5,4,6), (6,5,4), (5,5,5), (4,5,6), (5,6,4), (4,6,5)

Now sum the combinations for these valid cases:

(87)(64)(64)+(86)(64)(65)+(85)(64)(66)+(86)(65)(64)+(85)(65)(65)+(84)(65)(66)+(85)(66)(64)+(84)(66)(65)=11376\begin{aligned} &\binom{8}{7}\binom{6}{4}\binom{6}{4} + \binom{8}{6}\binom{6}{4}\binom{6}{5} + \binom{8}{5}\binom{6}{4}\binom{6}{6} \\ &+ \binom{8}{6}\binom{6}{5}\binom{6}{4} + \binom{8}{5}\binom{6}{5}\binom{6}{5} + \binom{8}{4}\binom{6}{5}\binom{6}{6} \\ &+ \binom{8}{5}\binom{6}{6}\binom{6}{4} + \binom{8}{4}\binom{6}{6}\binom{6}{5} = 11376 \end{aligned}

Therefore, the required numerical value is 1137611376.

Common mistakes

  • Counting only the number of distributions of 1515 questions among the three sections and stopping there is incorrect. The triples such as (7,4,4)(7,4,4) or (5,5,5)(5,5,5) tell only how many questions are taken from each section, not which actual questions are chosen. After finding valid triples, multiply by the corresponding combinations from each section.

  • Ignoring the upper bounds xA≤8x_A \le 8, xB≤6x_B \le 6, and xC≤6x_C \le 6 leads to invalid cases. The transformation to non-negative variables gives candidate solutions, but these must still be checked against the number of available questions in each section.

  • Using permutations instead of combinations is wrong because the order in which questions are selected does not matter. The correct counting is with (nr)\binom{n}{r} for each section, not with factorial arrangements.

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