Assistance w a math question please!

Test Preparation SAT Preparation
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<p>Hello, I was doing test 3 from the blue book and got stuck on a question. Section 2, question 18. This is the only question that I couldn't figure out and I have no idea what is the "key" word that I'm not seeing to figure out what topic knowledge I should have applied. I left out the question and still got 800 but I will like to understand the question and method behind it. </p>

<p>Question: Any 2 points determine a line. If there are 6 points in a plane, no 3 of which lie on the same line, how many lines are determined by pairs of these 6 points?</p>

<p>Any help will be greatly appreciated, thanks in advance!</p>

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<p>this is combinations. you just have to find how many different pairs you can make with 6 points. should be 6 ncr 2 which equals 15.</p>

<p>PS: you didn’t get an 800 on math. Very rarely (like 2 times EVER) has there been a math section where -1 is an 800. The score that the BB gives you is your RANGE of scores, and gives you what you are most likely to score when you actually take the SAT</p>

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<p>Thanks for the assistance but if you went to page 561 of the blue book (college board book) you will see that a raw score of 53 for this SPECIFIC (test 3) test is 800. This one doesn’t have a range like the others. Thanks :)</p>

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<p>Alternately:
If you are not familiar with combination/permutations just sketch 6 points on a circle (to avoid having three points on the same line) and methodically count all unique segments - connect one point with the rest, then move to the next point, and so on.
5+4+3+2+1 = 15.</p>

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<p>Ohhh, I love that idea thanks. I am familiar w it but I just think SATs want us to use a simple method w these questions and not “advanced math”. So when I see a triangle question I don’t use my sine, cos, trig rules as it’s not required for the test. Only problem is figuring out that simple method so you could finish fast. :confused: So I’m just looking for key words to direct me to a topic that is required for SAT math to get a simple/basic method. Thanks again!</p>

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<p>I prefer combinatorics to simply counting them all. Less likely to make a mistake if you know what you’re doing. Plus it’s not too advanced for the SAT.</p>

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<p>

That’s exactly what I said. If you don’t know how to use combinatorics - simply “count them all”. </p>

<p>Another way to answer the OP question would be 6x5/2 (which is 6C2, of course):
For each of 6 points there are 5 others to connect with, thus 6x5 connecting segments.
(You don’t need to know that this method is called the Fundamental Counting Principle in order to apply it.)
Each segment is counted twice (for example, AB and BA), so in order to eliminate duplicates 6x5/2 = 15.</p>

<p>How many diagonals are in octagon?
8C2-8=28-8=20 requires thinking;
methodical counting 5+5+4+3+2+1=20 is simple and not much longer.
Yet another way:
8x5/2=20 (for each of 8 start-points there are 5 end-points; duplicate segments must be eliminated).</p>

<p>It’s not the matter whether combinatorics is too advanced or not for he SAT.
SAT/ACT questions are designed in such a way that they can be answered either by counting or multiplication.</p>

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<p>I decided that I’d use combinatorics as I understand it instead of trying to find something simpler.</p>

<p>I have another question it’s a permutation. I got the answer but I wanna know if my thought process was right.</p>

<p>Question: Meredith has a red hat, a blue hat and a white hat. She also has three sweaters- one red, one blue and one white- and three pairs of jeans- one red, one blue and one white. Meredith wants to wear a red, white and blue outfit consisting of one hat, one sweater and one pairs of jeans. How many different possibilities does she have? </p>

<p>I disregarded the first sentence and started from “Meredith wants to wear a red, white and blue outfit…” I looked at it as the colors being most important, if they weren’t she’d just wear an all red, all blue, all white outfit or whatever color she felt like when it was time to choose an outfit. So the colors can’t be repeated. </p>

<p>So I said she had 3 choices for colors and 3 choices of clothing pieces (pair of jeans, sweater and hat) . If she didn’t the 3 pieces wouldn’t matter she would wear a sweater and pair of jeans w/o a hat. The clothing pieces can’t be repeated as well, it would be absurd for her to wear 3 jeans.
So 3!/(3-3)!</p>

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<p>Simple counting:
Colors: B - blue, R - red, W - white.
Items: 1 - hat, 2 - sweater, 3 - jeans.
Possibilities:</p>

<h2>1 2 3</h2>

<p>B R W
B W R
R W B
R B W
W B R
W R B
Six altogether.</p>

<p>Or:
3 color choices for item 1; 2 - for item 2 (not to repeat the color); 1 remaining color for item 3.
3x2x1 = 6.
Look Ma! No combinatorics! :D</p>

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<p>Thanks, I prefer the last part. Much easier.</p>

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<p>

</p>

<p>And the ETS writers have historically been very generous with the people who simply count the possible solutions. As far as I know, the method that ignores the calculator or the combination has repeatedly shown to be both idiot-proof and … faster, especially when one recognizes that the longer problems tend to have a short solution. </p>

<p>I have always felt that the ETS trap in such problems is set for people who will make mistakes in calculating the combinations and mess up between combinations and permutations.</p>

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<p>I reviewed permutations and combinations and in my opinion I find it much easier knowing whether it was a permutation or combination as well as knowing if there was a repetition or not. I also did realize that the wordy questions had a simple short solution like the one gcf101 assisted me with. I think the extra words were just there to distract. While I do think it’s possible to count, for me it’s too long and frustrating for a level 4 or 5 question knowing that I have limited time. </p>

<p>Know any SAT perm/comb questions that I could practice?</p>

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<p>What I would do is search the forum for words that typically appear in such questions.</p>

<p>Here’s an example on the word “arrange”</p>

<p><a href=“http://talk.collegeconfidential.com/sat-preparation/1547534-math-question-once-again.html?highlight=arrange[/url]”>http://talk.collegeconfidential.com/sat-preparation/1547534-math-question-once-again.html?highlight=arrange&lt;/a&gt;&lt;/p&gt;

<p>Another option is to peruse the Blue and Red Books or a stack of QAS. The problems are pretty common.</p>

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<p>Or try this in Google:</p>

<p>“how many ways” site:talk.collegeconfidential.com</p>

<p>A while back, I was on a tear writing a bunch of these. This search picked up a lot of them (and others).</p>

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<p>Thanks a lot everybody, greatly appreciated!</p>

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<p>I was going through some questions and I saw one with a repetition for a perm. Are there any combination questions w repetition for SATs?</p>

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<p>These types of combinatorics/ “all possible outcome” problems can be usually done two ways on the SAT’s. </p>

<p>For example, if the questions asked something like, “If there are 5 teams, and if each team plays exactly one game with every other team, how many total games are played?” you can use the two methods to get the answer. (The question you asked is no different than the question I just stated).</p>

<p>Approach 1: Use combinations. In order to have a match, we need to make 2 selections. I can pick any of the five teams as my first selection, and any of the remaining four as my second selection. So far I have 5*4= 20 matches. However, the order doesn’t matter, since team A playing team B is the same as team B playing team A, so we have to divide by 2 to get 10 total matches.</p>

<p>Approach 2: We can systematically count. First team must play 4 games (since it cannot play itself). Second team will play 3 games (since it has already played with the first team), etc. so the answer is 4+3+2+1=10.</p>

<p>In context of the original question, to define a line we need to make 2 selections. Any of the six points can be selected first, and any of the remaining five can be selected second. Since AB is the same line as BA, simply divide by 2. I’m sure you can now apply the second method to this question.</p>

<p>Ask yourself these three questions and you’ll never get a combinatorics (SAT level, at least) wrong:</p>

<ol>
<li>How many selections am I making?</li>
<li>How many objects am I picking from?</li>
<li>Does order matter?</li>
</ol>

<p>The distinction between combinations and permutations is dependent upon the third questions. If order matters, its called permutations; and if order doesn’t matter (in which case, you’ll be over counting so you must divide by something) its called combinations.</p>