<p>At a dance competition, each of six couples must compete against the other five couples in a dance-off three times before the winning couple can be declared. How many such dance-offs will occur?</p>
<p>A 12
B 33
C 45
D 60
E 63</p>
<p>I know what the answer is and how i got it, but i just need thorough explaining. Thanks</p>
<p>3 x 6! ?</p>
<p>1st couple will dance off against the remaining 5 three times = 5 x 3 = 15 dance offs
2nd couple will dance off against the remaining 4 three times = 4 x 3 = 12 dance offs
3rd couple will dance off against the remaining 3 three times = 3 x 3 = 9 dance offs
4th couple will dance off against the remaining 2 three times = 2 x 3 = 6 dance offs
5th couple will dance off against the remaining 1 three times = 1 x 3 = 3 dance offs</p>
<p>15+12+9+6+3=45</p>
<p>IDK if that’s the right answer</p>
<h2>1 2 3 4 5 6</h2>
<p>2 3 4 5 6
3 4 5 6
4 5 6
5 6
6</p>
<p>The numbers above the dashes represent couples 1 - 6. Each couple dances with the other 5 once for each round.</p>
<p>Count up all the numbers below dashes and multiple by 3 (for three rounds) to end up with 45.</p>
<p>Algebraically, I was always too lazy to figure this out. This always seemed faster.</p>
<p>6 * 5 = 30 permutations
30/2 = 15 combinations</p>
<p>Then it’s multiplied by 3 due to the context of the problem. 15 * 3 = 45.
simple as that.</p>
<p>187’s way is correct.</p>