CSAT Solved Papers/ 2023/Q4

2023 CSAT — Q4

Quant Counting & combinatorics 2.5 marks Medium

Raj has ten pairs of red, nine pairs of white and eight pairs of black shoes in a box. If he randomly picks shoes one by one (without replacement) from the box to get a red pair of shoes to wear, what is the maximum number of attempts he has to make?

  1. A 27
  2. B 36
  3. C 44
  4. D 45 Answer

Worked rationale

“Maximum number of attempts” means the worst case — the unluckiest possible draw before a wearable red pair (one left-foot red and one right-foot red) is forced.

Count the shoes:

  • Red: 1010 pairs =20= 20 shoes (1010 left ++ 1010 right).
  • White: 99 pairs =18= 18 shoes.
  • Black: 88 pairs =16= 16 shoes.

The adversary delays a red pair as long as possible:

  1. Draw all non-red shoes first: 18+16=3418 + 16 = 34 shoes — still no red.
  2. Then draw all 1010 red shoes of one foot (say all 1010 left-reds): 34+10=4434 + 10 = 44 shoes — now he holds 1010 left-red shoes but still no matched pair.
  3. The very next shoe — the 45th45^{\text{th}} — must be a right-red (the only shoes left are red right-foot), which completes a red pair.

So the worst case is 34+10+1=4534 + 10 + 1 = 45 attempts.

Answer: (d) 45.

Visual solution

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

Hand-drawn worked solution for UPSC 2023 CSAT Q4 — Counting & combinatorics
Tap the drawing to open it full size for the fine detail.

Why the other options miss

  • A
    counted pairs, not shoes: works in pairs rather than individual shoes, or stops at “any two red” rather than a matched left+right pair.
  • B
    cleared only one non-red colour: exhausts just one colour of the non-reds before grabbing reds, forgetting the second non-red colour must also be cleared in the worst case.
  • C
    stopped one shoe early: at 4444 he holds 1010 same-foot reds but no matched pair yet; the pair is forced only on the very next draw.

Specialist insight

The decisive subtlety is that a “red pair to wear” is a matched pair (one left, one right), not just two red shoes. That is why drawing all 1010 reds of a single foot (1010 shoes) does not yet give a pair — the worst case must absorb an entire foot’s worth of reds before the match is forced. Treating “red pair” as “any two reds” gives 34+2=3634 + 2 = 36 and loses the mark. Worst-case pigeonhole: clear every way to avoid the goal, then add one.

The trap, in one line

A wearable pair is left ++ right, so the worst case is all 3434 non-reds ++ all 1010 reds of one foot +1=45+ 1 = 45, not 3636.

← All 2023 CSAT questions