<p>The total number of ways to arrange 6 \(A\)'s and 4 \(B\)'s in a row is \(\dfrac{10!}{6! \times 4!} = {}^{10}C_4\). What is the total number of ways?</p>
Step-by-Step Solution
Key Concept: When arranging identical objects of multiple types, the total arrangements = total positions choose positions for one type (since remaining positions automatically get the other type). This is a fundamental principle of combinations with repetition of identical items.
<p><strong>Step 1:</strong> We have 10 total positions to fill with 6 identical A's and 4 identical B's.</p><p><strong>Step 2:</strong> The problem reduces to: choose 4 positions (or 6 positions) out of 10 for the B's (or A's). Remaining positions automatically get filled with A's (or B's).</p><p><strong>Step 3:</strong> Number of ways = <sup>10</sup>C₄ = 10!/(6! × 4!)</p><p><strong>Step 4:</strong> Calculate: <sup>10</sup>C₄ = (10 × 9 × 8 × 7)/(4 × 3 × 2 × 1) = 5040/24 = <strong>210</strong></p><p>∴ Answer: <strong>210</strong> (Option B)</p>
Correct Answer: B