# math question - combinations

<p>At a basketball tournament involving 8 teams, each team played 4 games with each of the other teams. How many games were played at this tournament. </p>

<p>A. 64
B. 98
C. 112
D. 128
E. 224</p>

<p>The answer is D. Could someone explain to me in detail how to solve this question?</p>

<p>Baffles me. I've tried it thrice, but I get C over and over. E is simply 8<em>7</em>4, an obvious trap. D is 2^7, but why, I can't understand.</p>

<p>I applied 4(7+6+5+4+3+2+1) to get C, and the rationale makes sense to me... Would like to know the solution too.</p>

Number of possibe matchups= 8 choose 2=28
Each matchup is played 4 times so total number of games=28*4=112</p>

<p>I'll explain briefly how you come up with the answer.</p>

<p>How many different combinations of 2, can be found in a group of 8? I won't trouble you with the thought. Simply bring your FX calculator, and press '8', followed by 'Shift' and then the 'divide' sign on the calculator (you'll notice that there's a letter 'C' under it), so now your calculator has the following written on its screen "8C". Good. How many combinations do we want in 8 again? 2. OK, so press 2 on the calculator. You now have "8C2". Press Equal. It gives you a total of 28 combinations. Since each 2 teams played a total of 4 matches, and not 1, we'll be multiplying 28 by 4 to give the correct answer of 112.</p>

<p>On a TI84/83 you type the number of teams (8) then go MATH>PRB>nCr>ENTER then type how many you're choosing at a time (2). You should get 28. 28*4=112. You could also add 7+6+5+4+3+2+1 then multiply by 4 if you write out combinations.</p>

<p>Here is Kaplan's explanation: If each team played 4 games with each of the other teams, then each team played (8)(4) = 32 games. Use this number to determine how many total games were played.</p>