CSAT Solved Papers/ 2022/Q27

2022 CSAT — Q27

Quant Counting & combinatorics 2.5 marks Easy

$A, B$ and $C$ are three places such that there are three different roads from $A$ to $B$ , four different roads from $B$ to $C$ and three different roads from $A$ to $C$ . In how many different ways can one travel from $A$ to $C$ using these roads?

A 10
B 13
C 15 Answer
D 36

Worked rationale

Two disjoint ways to get from $A$ to $C$ :

Via $B$ : choose an $A$ – $B$ road, then a $B$ – $C$ road — multiplication rule: $3 \times 4 = 12$ .
Direct $A$ – $C$ : $3$ roads.

These routes are mutually exclusive, so add (addition rule):

12 + 3 = 15.

Answer: (c) 15.

Visual solution

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

Hand-drawn worked solution for UPSC 2022 CSAT Q27 — Counting & combinatorics — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

added the via- $B$ legs: mis-handles the via- $B$ count as $3+4$ then $+3$ , instead of multiplying the two legs.
B

added every road count: sums all the roads $3 + 4 + 3$ , or wrongly nets $12+1$ , instead of multiplying the sequential legs.
D

multiplied the direct route in: multiplies everything ( $3 \times 4 \times 3$ ), folding the direct route into the product instead of adding it as a separate case.

Specialist insight

Multiplication for “and” (sequential legs $A\to B$ and $B\to C$ ), addition for “or” (via- $B$ or direct). The direct $A$ – $C$ roads are a separate case that must be added, not folded into the product — the trap (d) multiplies them in. $3\times 4 + 3 = 15$ .

The trap, in one line

Via $B$ : $3\times 4 = 12$ ; direct: $+3$ ; total $15$ (add the direct route, don't multiply it) $\Rightarrow$ (c).

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