Permutations & Combinations
Arrangements with identical objects
Grade 11

Question:

<p>The total number of flags with three horizontal strips in order, which can be formed using 2 identical red, 2 identical green, and 2 identical white strips, is equal to</p>
<p>\(4!\)</p>
<p>\(3 \times (4!)\)</p>
<p>\(2 \times (4!)\)</p>
<p>none of these</p>

Step-by-Step Solution

Key Concept: Since we have identical strips of each color, we need to count distributions of colors across three positions where the order of strips matters but identical colors are indistinguishable. This requires partitioning 2 red, 2 green, and 2 white strips into 3 positions.
<p><strong>Step 1:</strong> Understand the problem: We select 3 strips total from our collection (2 red, 2 green, 2 white) and arrange them in order (3 positions).</p><p><strong>Step 2:</strong> Each strip in each position must be one of three colors. Let's denote a flag by (c₁, c₂, c₃) where cᵢ ∈ {R, G, W}.</p><p><strong>Step 3:</strong> We need combinations where we use at most 2 of each color across 3 positions.</p><p><strong>Step 4:</strong> Total unrestricted arrangements = 3³ = 27</p><p><strong>Step 5:</strong> Subtract invalid cases (using more than 2 of any color):</p><p>• All 3 red: RRR (1 way)</p><p>• All 3 green: GGG (1 way)</p><p>• All 3 white: WWW (1 way)</p><p><strong>Step 6:</strong> Valid flags = 27 - 3 = <strong>24</strong></p><p>Alternatively, systematically: (R,R,G), (R,R,W), (R,G,G), (R,G,W), (R,W,W), (G,G,W), (G,W,W) with their permutations:</p><p>• 3!/(2!×1!) = 3 arrangements each for 6 types = 18</p><p>• 3! = 6 arrangements for (R,G,W) type = 6</p><p>• Total = 18 + 6 = 24</p><p>∴ Answer: B (24)</p>
Correct Answer: B

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