alleryFind the number of ways a person can travel from city A to city C if there are 3 routes from A to B and 2 routes from B to C.
A$3 + 2 = 5$
B$3 \cdot 2 = 6$
C$3!/2! = 3$
D$\binom{3}{2} = 3$
Answer & Solution
Correct answer: B. $3 \cdot 2 = 6$
1. This is a classic application of the MULTIPLICATION (counting) PRINCIPLE.
2. To complete the trip A → B → C, you must make BOTH choices: a route from A to B AND a route from B to C.
3. Independent choices: $3$ ways for first leg, $2$ ways for second.
4. Total ways: $3 \cdot 2 = 6$.
5. ADDITION applies when the EVENTS are alternatives (do X or do Y); MULTIPLICATION applies when both events must happen sequentially.
6. Option A uses addition (would count if you could only do one leg). Options C and D have no relevance.
_Source: syllabus Mathematics, Ch 6, §6.2 (Fundamental Principle of Counting), p. 2–4._
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