<p>There are <em>n</em> married couples at a party. Each person shakes hand with every person other than her or his spouse. Find the total number of hand-shakes.</p>
Step-by-Step Solution
Key Concept: Each person shakes hands with (2n-2) people (everyone except themselves and their spouse), giving n(2n-2) ordered pairs. Since each handshake involves two people, divide by 2 to get the actual count.
<p><strong>Step 1:</strong> Count total people: 2n people at the party (n couples).</p><p><strong>Step 2:</strong> Each person shakes hands with everyone except themselves and their spouse, so each person shakes hands with (2n - 2) people.</p><p><strong>Step 3:</strong> If we multiply n × (2n - 2), we get the total count treating each handshake as ordered (A→B and B→A counted separately).</p><p><strong>Step 4:</strong> Since each handshake is counted twice in the above product, divide by 2:</p><p style='text-align: center; font-weight: bold;'>Total handshakes = n(2n - 2)/2 = 2n(n - 1)/2 = <u>n(2n - 2)</u> = <u>2n(n - 1)</u></p><p><strong>Verification:</strong> For n=1: 2(1)(0) = 0 ✓ (one couple, no handshakes)</p><p>For n=2: 2(2)(1) = 4 ✓ (person A shakes with C and D; person B shakes with C and D = 4 total)</p><p>∴ Answer: <strong>2n(n-1)</strong></p>
Correct Answer: 2n(n-1)