Q5·CSAT · Prelims 2023

Ways to score exactly 25 runs

NumericalPermutations & CombinationsNumber theory● Hard

Question

In how many ways can a batsman score exactly 25 runs by scoring single runs, fours and sixes only, irrespective of the sequence of scoring shots?

Options

Answer

Explanation

Let the number of singles, fours, and sixes be x, y, and z respectively. We must find non-negative integer solutions to the equation: ⟨MATH⟩1x + 4y + 6z = 25⟨/MATH⟩.

Since x can always absorb any remaining deficit as single runs, the number of combinations depends entirely on the valid pairs of (y, z) such that 4y + 6z \le 25. Systematically test values for the largest multiplier, z:

If ⟨MATH⟩z = 4⟨/MATH⟩: 6(4) = 24. Remaining = 25 - 24 = 1. Max 4y \le 1 \implies y = 0. (1 solution: {0, 4})

If ⟨MATH⟩z = 3⟨/MATH⟩: 6(3) = 18. Remaining = 25 - 18 = 7. 4y \le 7 \implies y can be 0, 1. (2 solutions: {0, 3}, {1, 3})

If ⟨MATH⟩z = 2⟨/MATH⟩: 6(2) = 12. Remaining = 25 - 12 = 13. 4y \le 13 \implies y can be 0, 1, 2, 3. (4 solutions)

If ⟨MATH⟩z = 1⟨/MATH⟩: 6(1) = 6. Remaining = 25 - 6 = 19. 4y \le 19 \implies y can be 0, 1, 2, 3, 4. (5 solutions)

If ⟨MATH⟩z = 0⟨/MATH⟩: 6(0) = 0. Remaining = 25 - 0 = 25. 4y \le 25 \implies y can be 0, 1, 2, 3, 4, 5, 6. (7 solutions)

Summing all valid combinations: 1 + 2 + 4 + 5 + 7 = 19 ways.

To find combinations for Diophantine linear expressions with a generic slack variable (like single runs), bound and list cases using the largest operational coefficient.

Answer: (b).

Question details

Year

2023

Paper

CSAT

Question

Section

Numerical Ability

Sub-topic

Permutations & Combinations

Type