CSAT Solved Papers/ 2022/Q77

2022 CSAT — Q77

Quant Counting & combinatorics 2.5 marks Hard

There are 99 cups placed on a table arranged in equal number of rows and columns out of which 66 cups contain coffee and 33 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee?

  1. A 18
  2. B 27
  3. C 54
  4. D 81 Answer

Worked rationale

“Equal rows and columns” with 99 cups means a 3×33 \times 3 grid. Identical cups, so an arrangement is just a choice of which 33 of the 99 cells hold tea (the rest hold coffee).

Total placements of the 33 tea cups: (93)=84\binom{9}{3} = 84.

Complementary count — subtract those failing “every row has at least one coffee.” A row fails only if all 33 of its cells are tea (no coffee). Since there are exactly 33 tea cups, a failing row uses all of them, so at most one row can fail, and the number of such bad placements is

3 (choice of the all-tea row)×1=3.3 \text{ (choice of the all-tea row)} \times 1 = 3.

Valid arrangements:

84−3=81.84 - 3 = 81.

Answer: (d) 81.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2022 CSAT Q77 — Counting & combinatorics
Tap the drawing to open it full size for the fine detail.

Why the other options miss

  • A
    over-restricted the rows: forces something like one tea per row, drastically undercounting.
  • B
    wrong model: counts 333^3 (one tea slot per row, chosen independently), which is not the setup here.
  • C
    subtracted too much: removes column-failures as well (or applies a doubled correction to (93)\binom{9}{3}), over-deducting.

Specialist insight

Complementary counting is the clean route: total (93)=84\binom{9}{3} = 84 minus the only way a row can lack coffee — all three tea cups falling in one row (33 placements). No inclusion–exclusion overlap arises because three tea cups cannot empty two rows of coffee. 84−3=8184 - 3 = 81.

The trap, in one line

(93)=84\binom{9}{3} = 84 minus the 33 all-tea-row placements =81⇒= 81 \Rightarrow (d).

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