2023 CSAT — Q5

Quant Counting & combinatorics 2.5 marks Medium

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?

A 18
B 19 Answer
C 20
D 21

Worked rationale

Let $a$ singles, $b$ fours, $c$ sixes, all non-negative integers, with

a + 4b + 6c = 25.

Since order is irrelevant, count $(a,b,c)$ triples. Once $b,c$ are fixed, $a = 25 - 4b - 6c$ is forced (and must be $\ge 0$ ), so just count the admissible $(b,c)$ :

$c = 0$ : $4b \le 25 \Rightarrow b = 0,1,2,3,4,5,6$ — $7$
$c = 1$ : $4b \le 19 \Rightarrow b = 0,1,2,3,4$ — $5$
$c = 2$ : $4b \le 13 \Rightarrow b = 0,1,2,3$ — $4$
$c = 3$ : $4b \le 7 \Rightarrow b = 0,1$ — $2$
$c = 4$ : $6 \cdot 4 = 24,\ 4b \le 1 \Rightarrow b = 0$ — $1$

( $c = 5$ gives $30 > 25$ , stop.) Total $= 7 + 5 + 4 + 2 + 1 = 19$ .

Answer: (b) 19.

Why the other options miss

A

one boundary case lost: drops an edge case (typically the all-singles $a=25$ solution at $b=c=0$ , or the tight $c=4$ case).
C

an invalid case counted: admits a non-integer or negative $a$ in one $(b,c)$ slot, or double-counts a case.
D

counted the order of shots: treats the shot sequences as distinct, or mis-bounds $b$ , inflating the total beyond the partition count.

Specialist insight

The move that scores is to eliminate the largest-value coin first (the six), looping $c$ from its maximum down, because $c$ has the fewest cases. With $a$ determined by $b,c$ , the problem collapses to counting integer points under $4b + 6c \le 25$ — a two-variable bounded sweep finished in well under two minutes. Counting $a$ as a free third variable, or attaching order to the shots, is the time-and-mark sink.

The trap, in one line

$a$ is forced once $(b,c)$ are chosen, so just count $(b,c)$ with $4b+6c\le 25$ — five $c$ -cases give $7+5+4+2+1=19$ .

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