CSAT Solved Papers/ 2023/Q65

2023 CSAT — Q65

Quant Counting & combinatorics 2.5 marks Medium

In an examination, the maximum marks for each of the four papers namely $P, Q, R$ and $S$ are $100$ . Marks scored by the students are in integers. A student can score $99\%$ in $n$ different ways. What is the value of $n$ ?

A 16
B 17
C 23
D 35 Answer

Worked rationale

$99\%$ of $400$ total marks is $396$ . Instead of counting score-tuples summing to $396$ , count the deficit from full marks: the total lost is $400 - 396 = 4$ , distributed over the four papers, with each paper losing $0$ to $100$ marks.

Let $x_P, x_Q, x_R, x_S$ be the marks lost, $x_i \ge 0$ and $x_P + x_Q + x_R + x_S = 4$ . Since the cap $x_i \le 100$ is never binding (the total deficit is only $4$ ), this is a plain stars-and-bars count:

n = \binom{4 + 4 - 1}{4 - 1} = \binom{7}{3} = 35.

Answer: (d) 35.

Why the other options miss

A

wrong formula: uses $\binom{6}{2} = 15$ (deficit $3$ ) or a $2^4$ -style count, applying the wrong stars-and-bars parameters.
B

off by one: computes $\binom{7}{3}$ -adjacent values or adds the deficit wrongly ( $396$ vs $400$ ), landing near but below $35$ .
C

corrected for a case that doesn’t exist: subtracts phantom “over- $100$ ” cases that don’t exist (the cap is never reached for a deficit of $4$ ), undercounting.

Specialist insight

The move that scores is the deficit reframe: counting ways to lose $4$ marks is far cleaner than ways to score $396$ , and it makes the $0 \le x_i \le 100$ cap obviously non-binding (you can’t lose more than $4$ overall). Then $\binom{n+k-1}{k-1}$ with $n=4$ stars, $k=4$ bins gives $\binom{7}{3} = 35$ . Trying to apply the upper-bound (inclusion–exclusion) correction is wasted work here — and the source of the $23$ trap.

The trap, in one line

Count the deficit: $x_P+x_Q+x_R+x_S = 400-396 = 4$ , caps non-binding $\Rightarrow \binom{7}{3} = 35$ $=$ (d).

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