Permutations & Combinations
Symmetric arrangements
Grade 11
Question:
<p>\(2m\) white counters and \(2n\) red counters are arranged in a straight line with \((m+n)\) counters on each side of a central mark. The number of ways of arranging the counters, so that the arrangements are symmetrical with respect to the central mark is</p>
<p>\({}^{m+n}C_m\)</p>
<p>\({}^{2m+2n}C_{2m}\)</p>
<p>\(\dfrac{1}{2}\dfrac{(m+n)!}{m!n!}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: For symmetry about the central mark, once you place counters on one side (m+n positions), the other side is completely determined. You need to distribute m white and n red counters into m+n positions on one side, and the remaining m white and n red automatically go to the other side in mirror positions.
<p><strong>Step 1:</strong> Understand the setup. Total: 2m white + 2n red counters arranged in a line with (m+n) counters on each side of a central mark. This means m+n positions on the left, central mark, m+n positions on the right.</p><p><strong>Step 2:</strong> For symmetry about the central mark: if position i on the left has a white counter, position i on the right must also have a white counter (in mirror position).</p><p><strong>Step 3:</strong> Therefore, we only need to arrange m white and n red counters in the (m+n) positions on ONE side. The other side is automatically determined by symmetry.</p><p><strong>Step 4:</strong> The number of ways to choose m positions out of (m+n) positions for white counters (with remaining n positions for red) is:</p><p>$$\binom{m+n}{m} = \binom{m+n}{n} = \frac{(m+n)!}{m! \cdot n!}$$</p><p><strong>Step 5:</strong> This accounts for all symmetrical arrangements since fixing one side completely determines the symmetric arrangement.</p><p>∴ Answer: A</p>
Correct Answer: A