CSAT Solved Papers/ 2023/Q66

2023 CSAT — Q66

Quant Counting & combinatorics 2.5 marks Easy

A flag has to be designed with $4$ horizontal stripes using some or all of the colours red, green and yellow. What is the number of different ways in which this can be done so that no two adjacent stripes have the same colour?

A 12
B 18
C 24 Answer
D 36

Worked rationale

Colour the stripes top to bottom. The first stripe is free; each later stripe must differ only from the one directly above it.

Stripe $1$ : $3$ choices (red, green, yellow).
Stripe $2$ : $2$ choices (anything but stripe $1$ ).
Stripe $3$ : $2$ choices (anything but stripe $2$ ).
Stripe $4$ : $2$ choices (anything but stripe $3$ ).

By the multiplication principle:

3 \times 2 \times 2 \times 2 = 24.

Answer: (c) 24.

Why the other options miss

A

an arithmetic slip: drops one factor of $2$ (e.g. $3\times 2\times 2 = 12$ , colouring only three stripes).
B

wrong formula: uses $3\times 3\times 2$ or a wrong per-stripe count, mis-applying the “differs from neighbour” rule.
D

missed a case: over-counts by allowing $3$ choices for a stripe that must avoid its neighbour ( $3\times 3\times 2\times 2$ ), forgetting each interior stripe has only $2$ valid colours.

Specialist insight

The standard “path-colouring” count: with $k$ colours and $m$ stripes in a line and the only rule being “adjacent differ,” the count is $k(k-1)^{m-1}$ . Here $k = 3$ , $m = 4$ , giving $3 \cdot 2^3 = 24$ . The phrase “some or all of the colours” is a non-constraint — it just means a colour may be unused; it does not add cases. Recognising the $k(k-1)^{m-1}$ template makes this a $30$ -second item.

The trap, in one line

$3$ for the first stripe, $2$ for each of the next three: $3\cdot 2^3 = 24$ ("some or all" adds nothing) $=$ (c).

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