Permutations & Combinations
Counting with Restrictions
Grade 11
Question:
<p>From 5 married couples (5 husbands and 5 wives), 2 husbands are selected for two different sides A and B of a game. Then their wives are excluded. From the remaining 3 wives, 2 wives are chosen, and they can interchange their sides. Find the total number of ways to form the two teams.</p>
Step-by-Step Solution
Key Concept: Apply the multiplication principle: first select the husbands, then select wives from the remaining pool, then count arrangements of wives to sides.
<p><strong>Step 1:</strong> Select 2 husbands out of 5 for sides A and B:</p><p><span class="math">\binom{5}{2} = 10 \text{ ways}\</span></p><p><strong>Step 2:</strong> The wives of these 2 husbands are excluded. Select 2 wives from the remaining 3 wives:</p><p><span class="math">\binom{3}{2} = 3 \text{ ways}\</span></p><p><strong>Step 3:</strong> The 2 wives can interchange their sides A and B:</p><p><span class="math">2! = 2 \text{ ways}\</span></p><p><strong>Step 4:</strong> By the multiplication principle:</p><p><span class="math">\text{Total ways} = 10 \times 3 \times 2 = 60\</span></p><p>∴ The required number of ways is <strong>60</strong>.</p>
Correct Answer: 60