A couple of Math SAT questions!

<p>These questions are from a recently administered SAT, and this just has the answers, no explanations, can someone explain these to me?!</p>

<ol>
<li><p>Each of the 7 basketball teams plays each of the other teams just once. How many games will there be?
A.17
B.18
C.19
D. 20
E. 21 - this is the answer</p></li>
<li><p>An alarm system uses a three-letter code and no letter can be used more than once. How many possible codes are there?
A. 13,824
B. 15,000
C. 15,600 - this is the answer
D. 17,000
E. 17, 576</p></li>
<li><p>How many seat arrangements can be made for five students in a row if one of them does not want to take the first seat?
A. 120
B. 96 - this is the answer
C. 85
D. 80
E. 75</p></li>
</ol>

<p>7) 6 + 5 + 4 + 3 + 2 + 1 = 21</p>

<p>Team 1 plays the 6 other teams. Team 2 plays the 5 remaining teams. Team 4 plays the 4 remaining teams. Team 3 plays the 3 remaining teams … see a pattern?</p>

<p>6 + 5 + 4 + 3 + 2 + 1 = 21 games total </p>

<p>12) “three-letter code” -> you can use the letters A-Z (26 letters in all)</p>

<p>“no letter can be used more than once” - once you use a letter, it’s out </p>

<p>Thus, the answer is:</p>

<p>26 * 25 * 24 = 15,600 possibilities.</p>

<p>27) There are 5 seats to be filled:</p>

<hr>

<p>But one student won’t take the first seat. So you can put any one of the 4 remaining students in the 1st seat.</p>

<p>You can also put 4 students in the second seat, 3 in the third seat, 2 in the fourth seat, and 1 in the last seat. </p>

<p>4 * 4 * 3 * 2 * 1 = 96</p>

<p>This may help you visualize the possibilities. The number underlined represents the number of people that you can put in the seat. </p>

<p>4 * 4 * 3 * 2 * 1</p>

<p>GREAT! thanks for this thread</p>

<p>I knew there was a simple way to solve problems like this, I usually would encounter it on the test and try all possibilities. Thankfully, you helped eliminate that dreadful routine.</p>

<p>anymore examples as to help solidify these solutions?</p>

<p>You can also think of the first question like the other two, with simple counting principle. There are 7 choices for the first team and 6 for the second for each game. The only thing is each game counts twice this way, because two teams play in each game, so there are 42 games experienced by a team, but each game is experienced twice.</p>