Posts tagged with: counting

The figure above represents four offices that will be
assigned randomly to four employees, one employee
per office. If Karen and Tina are two of the four
employees, what is the probability that each will be
assigned an office indicated with an X ?

The figure is Four boxes or (offices) and two of them are labeled X.

See a discussion of this question from the Tumblr Q&A here.

In a post I read of yours you said that with the exception of 3 things (lines, ……) there are no other formulas you should memorise. In other posts of your I have seen you use the combinations formula. What are your thoughts on using the permutations and combinations formula ?

Could you point me in the direction of any posts you have made (or section in your book) where you discuss counting problems specifically.

Thanks man

You never need permutations or combinations for the SAT. Often, trying to use them will lead to mistakes. My official position is that you don’t need them, which is why I leave them out of my book completely.

You can find some counting posts here:

Carmen arranges stacks of blocks so that each successive level has 1 fewer block then the level below it and the top level has 1 block. Carmen wants to make such a stack with 12 levels. how many blocks does she need?

The top level will have 1 block, then the next level down will have 2, then the next level down will have 3, etc. If she wants 12 levels, then we’ll have to add a block 12 times!

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78

Ten distinct lines lie in the same plane. No pair of these lines is parallel to each other, and no more than two of the lines intersect at any one point. How many points lie on more than one of these ten lines? A)10 B)36 C)45 D)55 E)100

This is so similar to a question I was recently asked on the Tumblr Q&A that it must be from the same source. When the question asks “how many points lie on more than one of these ten lines,” it’s just asking for the number of intersections. So follow the same logic I used to solve the other question:

  • Draw the first line. It’s all by itself, so it intersects nothing.
  • Draw the second line. It intersects the first line. That’s 1 intersection.
  • Draw the third line. It intersects both existing lines. That’s 2 more intersections.
  • Draw the fourth line. It intersects all three existing lines. That’s 3 more intersections…

…and just keep going. The first four lines give you 0 + 1 + 2 + 3 intersections. The next six will continue that pattern. The number of intersections of those ten lines will be:

0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

I’m now realizing that I’ve done this question before, too. Check out my other solution (same approach, different formatting) here.

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How’s everyone else doing on this quiz?

10% got 5 right
17% got 4 right
15% got 3 right
26% got 2 right
31% got 0 or 1 right

 

* I don’t advocate violence towards cats (or other animals). “There’s more than one way to skin a cat” is a phrase that I used to hear all the time growing up, but that I now realize (having received some mortified stares at its utterance) that it’s not as common as I thought it was. It just means that a problem might have more than one solution. I still say it even though I have to clarify it now because I’m stubborn. 

I chimed in on a thread at College Confidential recently about a probability problem that apparently came from Dr. Chung’s book. The first few respondents provided completely legit (but rather technical) explanations of the problem using nCr, and then someone asked whether there was another way. So I jumped in with the way I prefer to solve most counting and probability questions on the SAT: short lists. For the most part, all the solutions offered in the thread were valid.

Why am I posting about that conversation here? Because it underscores an important fact: there are often multiple ways to solve SAT math problems. That’s one of the beautiful things about math in general, actually: it’s built on itself. That’s why you learned addition before multiplication, multiplication before exponents, geometry before trigonometry, etc. My recollection of learning nCr techniques was that they were slowly introduced to us as general solutions to simple combination problems we could solve with simpler counting principles—the kinds of problems you’ll see on the SAT.

I get a rush out of breaking a problem down until it’s so easy a caveman could do it. That’s one reason I’m a pretty good SAT teacher. But you’re probably not aspiring to a career in test prep; you’re probably just trying to score high enough on the SAT to move on with your life.

So it’s completely up to you whether you solve a combination/permutation problem on the SAT simply, the way I like to, or with robust, scalable techniques that would work just as well with many more elements. But I like to remind my students that you’ll never need nCr on the SAT; the numbers will stay small. Just like you’ll never need trigonometry or logarithms, even though you might once in a while spot a problem that can be solved with them.

At the end of the day, I’m all about information. In my ideal world, you’ll know more than one way to solve every problem, so you’ll be able to make informed decisions on whether to, say, plug in or do the algebra on a question-by-question basis, instead of being forced into algebra on every question because it’s the only way you know.

You’ve got a math teacher in school 5 days a week who will advocate the mathy way. You’ve got me, if you want, to show you alternatives. Once you’ve poked around and seen what’s out there, you decide the level of complexity that’s most comfortable for you and leads to your best score.

Source.

One technique-able counting problem type that you might come across on the SAT is what I’ll call a “possibilities” problem*. It might involve cards (but not playing cards – the SAT doesn’t like those), or pictures being lined up on a wall in different orders. Your job will be to determine the number of possible outcomes given a particular scenario. Like so:

  1. Mike is arranging seven of his various awards and commendations on a shelf in his office. If he insists that his hard-fought Class of 1999 Math Award be placed in the center, in how many different orders could he arrange the seven items?
     
    (A) 60
    (B) 72
    (C) 120
    (D) 720
    (E) 1440

The best way to tackle a possibilities problem like this is to draw a bunch of blanks like you’re about to play Hangman, and then start thinking, methodically, through the choices you have at every step along the way. I’m going to illustrate this process with slightly more thoroughness than you probably will on test day (you won’t need to make up award names, but I will because it’s hilarious):

Position
1
2
3
4
5
6
7
Award
Class of 1999 Math Award
Choices
1

 

First, as I did above, you must account for any special conditions or restrictions. The Math Award must go in the middle, so there’s only one choice for Position 4.

Position
1
2
3
4
5
6
7
Award
Invisible Man Award
Class of 1999 Math Award
Choices
6
1

 

Once all the restrictions are accounted for, start filling in the rest of the spaces. To fill Position 1, Mike has 6 different awards to choose from. Say, for argument’s sake, he chooses the Invisible Man Award next.

Position
1
2
3
4
5
6
7
Award
Invisible Man Award
Acne League – Most Improved
Push Ups Contest – Last Place
Class of 1999 Math Award
Spin the Bottle – Luckiest Player Ever
5th Grade Science Fair – 2nd Place
Shortest Fight in School History – Loser
Choices
6
5
4
1
3
2
1

 

To fill the next spot, since he used up the Invisible Man Award, he has 5 awards left to choose from. Once he chooses the Acne League – Most Improved award for Position 2, he has 4 choices for Position 3. He continues this process until he’s filled all the positions.

To calculate the number of different arrangements Mike could have made, multiply the number of choices he had at every step:

6 × 5 × 4 × 1 × 3 × 2 × 1 = 720

When solving a possibilities problem, set up the hangman blanks, then imagine yourself actually performing the task described. First take care of special conditions or restrictions, and then take care of everything else. At every step, ask yourself “How many choices do I have?” And then ask yourself “How many choices do I have now?” And then – you guessed it – ask yourself how many choices you have again.

Stop when you run out of choices.

Possibly you’re interested in more practice?

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*Aside from this footnote, I’m deliberately avoiding the terms “permutation,” “combination,” and “factorial” here. That’s not because I don’t know them; it’s because I’ve found that they manufacture more confusion than they alleviate on the SAT. Remember: The SAT is not a math test! If you prefer to solve the questions laid out here by cramming them into nPr and nCr notations in your calculator, be my guest, but don’t cry to me when you miss counting questions on the SAT.

An unfortunate truth about the SAT is that while many questions can be answered with snappy tricks (many of which can be found on these pages), not all of them can. Most “counting” questions (and probability questions, for that matter) fall into this category.

Yes, I’m serious. Most.

Basically, if you don’t see within 15 seconds or so that you’re dealing with a matching problem, or a possibilities problem where you can just set up hangman blanks and count, then you should bail on looking for shortcuts and just start listing things. That’s right. List them.

Stop complaining. Listing them isn’t “the long way”! Sitting there with your leg shaking and your hands on your head trying to see a shortcut where there is no shortcut while time ticks away is “the long way”!

The long way is the short way

(Grid-in)

  1. How many positive integers less than 100 are not divisible by 7?

OK, so maybe I lied a little bit: there is a small shortcut, which is to treat this like a shaded region problem and find the opposite of what they’re asking for. In other words, list all the positive integers less than 100 that are divisible by 7:

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98

I’m counting 14 there. Since there are 99 total positive integers less than 100, the answer is 99 – 14 = 85. Nice, right?

I get a lot of questions from students about listing problems like these, and their questions always end the same way: “Is there a faster way?” Not really. But if you stop worrying about a faster way and just start listing as soon as you come across this kind of question, and you count carefully, you’ll probably get through it without much of a problem.

UPDATE: For this particular problem, there is one other shortcut. I hesitate to even mention it because I don’t want to dilute the point that your general reaction to a non-pattern-conforming counting question should just be to start listing, but when you get “How many positive integers less than m are not multiples of n?” questions you can also follow these steps:

  1. Find the greatest multiple of n that’s less than m. In this case, 98 is the greatest multiple of 7 that’s less than 100.
  2. Divide by n. In this case, 98/7 = 14.
  3. That means there are 14 multiples of 7 less than 100.

This Labor Day weekend I’m going to the wedding of one of my oldest friends from my hometown. Combine the fact that the number of my friends who aren’t married is dwindling dangerously close to zero and the fact that I wasn’t able to attend this year’s fantasy football draft so I had to autodraft and the fact that this weekend marks the end of the summer and OMG WHAT IS HAPPENING TO MY LIFE.

This weekend’s prize: you guessed it — free access to the Math Guide Beta. Make sure you don’t comment anonymously if you want to win.

In a certain youth soccer league, each team plays each other team exactly one time per season. If, during a certain season, there are 231 games played, how many teams were in the league that season?

First correct answer in the comments wins Beta access, and the undying adoration of my throngs of loyal readers.

UPDATE: Nice work, “AP FRQ Solutions” (whoever you are). I’ve shared the Beta document with you. Use it in good health.

Solution below the cut.
If you look back at my post about matching questions, you’ll notice a neat little tidbit hidden in there: if you’re looking for the total number of matches that can be made between a certain number of teams, you can take the number of teams, subtract 1, and take the sum of that number and each number lower than it to find the number of matches. In other words, if you have 7 teams, they will need 6+5+4+3+2+1 matches to each play each other once. Neat, huh?

So you can use that to solve this one quick and dirty:

1+2 = 3
1+2+3 = 6
1+2+3+4 = 10…

1+2+3+4+…+20+21 = 231.

So there must be 22 teams in the league.

Note: even if you didn’t know this little factoid, you might have been able to solve this way by finding the number of matches needed for smaller leagues, and looking for patterns in the number of matches required as the leagues get bigger.

If you’re looking for a mathier explanation, I think AP FRQ Solutions did a decent job in the comments.

The SAT will throw two common kinds of “counting” problems your way. I’ll handle one of them in this post. The other kind, well, I’ll get to it when I get to it. 🙂

I like to call this kind of problem a matching problem. It’ll usually involve a bunch of people who all need to shake hands, or a league in which every team needs to play every other team. Or a massage club where everyone has to give everyone else a back massage. I don’t know…whatever. Everyone has to touch everyone at the touching party?

…this is going nowhere. Let’s see an example.

  1. Each team in a kickball league plays each other team 4 times during the season. If there are 7 teams in the league, how many games long is the season?
     
    (A) 28
    (B) 56
    (C) 84
    (D) 112
    (E) 116


What you’re going to want to do here is draw a diagram.

Arrange the letters A-G (representing the 7 teams) in a large circle. Now draw
lines connecting each letter to each other letter, carefully counting as you draw. (If you try to count after you’re done drawing, you’re going to have a pretty difficult time getting an accurate count.) The best way to go about this is to draw every line that you can that originates at
A, and then do the same for B, etc.

You’ll know you’re done when you have something that resembles a star with all its outer points connected. Like so:

The number of lines you just drew – 21, you awesome counter you – equals the number of games required for each team to play each other once. If each team has to play each other 4 times, multiply 21 by 4 to get the answer: 84! BAM!

Pretty amazing, right? It’s just so…beautiful. No, stop crying. It’s totally inappropriate for you to be crying right now. I know it’s pretty but you need to stop. I refuse to move on until you stop crying.

Of course, if you want to represent the above diagram mathematically, you could say that Team A needs to play all the other teams, so it plays 6 games. Then Team B needs to play all the teams except Team A (since they already played), so that’s 5 more games. Follow that line of reasoning until its end and you get:

6 + 5 + 4 + 3 + 2 + 1 = 21

But I think the star is prettier (and easier to remember).

Care to try a few more?

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I lifted a box that was too heavy this week and I screwed up my back so bad that every time I put weight on my right foot, searing pain shoots up my entire right side. Getting older is awesome!

The prize this week for the first correct response: FREE Beta Access to my book. A $5 value! Read up on the deets, if this is the first you’re hearing of it.

How many positive integers less than 1,000 contain exactly one odd digit?

Put your answers in the comments, and I’ll post the solution and contact the winner (if there is one) on Monday. Good luck!

UPDATE: Congratulations to Chong Lee, who nailed it first. Welcome to the Beta, Chong Lee. I hope you enjoy the book.

Solution below the cut.

The best way to solve a question like this is to count possibilities for each digit. There are three places your single odd digit could appear: the hundreds digit, the tens digit, or the units digit. For each of those possibilities, you have to count all the possible positive outcomes.

If your odd digit is the hundreds digit, for example, then you count like this:

Hundreds digit is odd
  • There are 5 odd digits (1, 3, 5, 7, 9) that could occur as the hundreds digit.
  • There are 5 even digits (0, 2, 4, 6, 8) that could occur as the tens digit.
  • There are 5 even digits that could occur as the units digit.

5 × 5 × 5 = 125 possible outcomes in which the single odd digit is in the hundreds place.

What adds an extra degree of difficulty to this question is that the number of possible odd digits and the number of possible even digits is always the same. Sorry about that. That’s why it’s a challenge question.

Moving on, we still have to account for the odd digit coming in the tens or ones places.

Tens digit is odd
  • There are 5 even digits that could occur as the hundreds digit. (Remember, zero still counts. 010, or just plain old 10, has one odd digit.)
  • There are 5 odd digits that could occur as the tens digit.
  • There are 5 even digits that could occur as the units digit.

5 × 5 × 5 = 125 possible outcomes in which the single odd digit is in the tens place.

Ones digit is odd
  • There are 5 even digits that could occur as the hundreds digit.
  • There are 5 even digits that could occur as the tens digit.
  • There are 5 odd digits that could occur as the units digit.

5 × 5 × 5 = 125 possible outcomes in which the single odd digit is in the ones place.

How many positive integers less than 1000 contain exactly one odd digit?

125 + 125 + 125 = 375.

Magic, right? I know. If you want to see it done in a way that it could never be done in a test environment, I’ve confirmed the results in a spreadsheet for all the doubters. Click here.