kitchen table math, the sequel: The New Math SAT Game Plan

## Wednesday, March 2, 2011

### The New Math SAT Game Plan

The New Math SAT Game Plan is a fantastic book.

Amazing.

Last night I read the section on counting.

Counting has been a massive struggle for me. I worked my way through Dolciani's chapter on combinations and permutations, did all the exercises, and ended up pretty much where I started out: basically, being "counting" blind: all the problems look alike. I can't tell the difference amongst them, and I can't tell when and where I would do what or why. I have been utterly flummoxed. *

Last night, reading Keller, everything clicked.** From one moment to the next, I abruptly understood why all the problems look alike (to me) and what the solutions have in common logically. I finished up Keller's 5-page explanation and did all of his exercises quickly and correctly. Easy-peasy.

Later on, I'll try Dolciani's exercises and see how I fare.

Here's how Keller explains the counting principle:
Your favorite restaurant offers a combo-meal. You get to pick one each from a menu of 6 sandwiches, 4 side salads, 5 beverages, 10 desserts and 3 sounvenir toys. You decide to eat at this restaurant once every day, ordering a sandwich, salad, beverage, dessert and toy every time, until you have had every possible combination. To the nearest whole number, how many years will it take you??

Do NOT attempt to list all of the combinations. Instead, learn the Counting Principle:

In any situation where you are faced with a series of decisions keep asking yourself:

"Now, how many choices do I have?"

until the last decision has been made. Then to find the overall number of combinations, you multiply together all the numbers of choices you had for each decision.

So in the example I have given you, you have to choose your sandwich from 6 choices, then your salad from 4 choices, your beverage from 5, your dessert from 10 and your toy from 3. And then you are done making decisions. So you multiply and find that there are

6 x 4 x 5 x 10 x 3 =3600 combinations. And then we can divide by 365 days in a year and find that it would take just under 10 years to order every combination. That's a long time but don't be surprised. When you have lots of decisions, or lots of options you get big numbers.

The example I just gave you is one of the easiest kinds of counting problems you'll see on an SAT. Many of the other varieties are a little harder to recognize and a little trickier to answer, as the next few examples will show.

After this, Keller shows how the solution to the problem where you've got a family sitting in a row and the mom and dad have to sit on the end chairs while the four kids can sit on any of the in-between chairs (and how many combinations is that???!!!) follows exactly from the solution to the how-many-combinations-in-the-restaurant problem.

Fantastic!

I am SO happy to know how many ways a family of 6 can sit in 6 chairs with both parents occupying the end chairs.

Seriously.

Now how many choices do I have?

* I have yet to use either of the two resources you all left for me: the Arlington Agebra Project and a web page created by a math professor who may have sent me the link via email (I don't remember - !). Since I don't remember, I won't link here.

** Well...not everything. Still having trouble with the SAT counting problem that nearly did me in last summer - but I now understand the first part of the problem.

Catherine Johnson said...

OK, I spoke too soon.

Still a great book, though.

Niels Henrik Abel said...

That's pretty much how I approach it when I have to explain the basic counting rule:

How many "slots" do you have to fill? (Draw the slots on the board.)

How many choices are there for each slot? (Write the appropriate number in each slot.)

Put a multiplication sign in between each number, and an equal sign at the end. Multiply.

Permutations of n distinct objects taken r at a time can also be treated in this manner - e.g., if there are 100 participants in a contest, in how many ways can 1st, 2nd, and 3rd place prizes be awarded?

Daniel Ethier said...

That is exactly what I do with MATHCOUNTS problems. I can never remember the formulas and even if I could, the problems always have some twist that makes them useless.

But when you walk through everything and ask how many choices do I have now (you also need to ask whether order matters -- if not you need to divide by the number of ways to arrange things to eliminate double counting) it works much better.

Catherine Johnson said...

Hi Niels!

Now I'm having trouble figuring out combinations .... at least, I think that's what I'm having trouble with.

I can only easily solve the first part of this SAT problem:

Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

I get that there are 60 ways to choose 3 employees for the office.

But that's all.

I particularly don't understand this part of the solution:

From the ￼3 male workers, 2￼ can be chosen in 3￼ different ways.

I understand it when I write out the possibilities, but I don't understand it when I try to apply the "How many choices do I have?" approach.

When I ask myself how many choices I have left, I end up with 3x2.

Allison said...

I'm not yet answering your question, but trying to slow you down and start answering pieces of this problem first.

when people are assigned ways of seating in seats, then ORDER matters. There's a clear notion someone goes on the leftmost seat, someone goes on the next leftmost seat that remains, etc.

So, the assignment in order of
Catherine, Ed, C, A, J, and S
is different than
Ed Catherine C A J S

When you assign people to a room, order doesn't matter.

Are there still 60 ways to put 3 people in one room and 2 in another?

Saying "I chose Catherine, that leaves 4 choices for the second slot, I pick Ed, that leaves 3, I choose A"

you'll count "I chose Ed, that leaves 4 choices, I pick Catherine, that leaves 3, I choose A..." as two of your 60.

Allison said...

(read this comment later, after you've thought for a while about the above)

But let me add that you're supposed to be proficient enough with counting by this problem that you "See" another "trick", called the pigeonhole principle.

What if the problem had asked how many ways are there to assign 2 women to the cubicle? Or how many ways are there to assign two women and 1 man to the office? would you have started counting out the ways to deal with all 5 employees?

Michael Weiss said...

Let's do this in two parts.
Part 1 is: How many different ways of assigning people to the various workspaces are there, in total?
Part 2 is: How many different ways are there that meet the conditions specified by the problem?

Then we'll divide the result of Part 2 by the result of Part 1, to get a probability.

So, Part 1:
We need to choose 2 people who get cubicles. We have 5 choices for the first cubicle; then 4 choices for the second one. Once those choices are made, there are no more choices; the remaining 3 people all get offices. So in all there are 5*4 decisions to make. However, this is actually an overcount -- it treats "Jim in cubicle 1 and Paul in cubicle 2" as a different configuration from "Paul in cubicle 2 and Jim in cubicle 1". So we need to divide that result by 2. Therefore there are 10 ways of assigning people to workspaces.

Now for Part 2. We need to choose a man for one cubicle and a woman for the other. There are 3 choices for the man who gets a cubicle; then there are 2 choices for the woman who gets a cubicle. All together there are 3*2 choices. This time there is no overcount, because if "Bob in the male cubicle and Nancy in the female cubicle" is one configuration, you do NOT get another by switching the names around. So there are 6 configurations.

Synthesis: In 60% of the arrangements, the cubicles end up assigned to 1 man and 1 woman, with the offices going to the other 2 men and the other 1 woman.

Now for Part 2. We want to choose a man to the first office; there are 3 choices. Having done that, we want to choose another man for the second office; now there are only 2 choices. Having done that, we assign a woman to the last office: there are 2 choices. All together we have 3*2*2 choices. Again, this is an overcount; we need to divide by the number of rearrangements of 3 people, which (as before) is 3*2. So all together there are (3*2*2)/(3*2) ways of assigning the offices in accordance with the specified constraints. Having done so, the remaining man and woman get the cubicles automatically, so no more choices get made. There are thus 2 ways of doing this. (In hindsight, those 2 choices amount to

Allison said...

(btw, my first comment is supposed to lead you to answer why the "how many choices I have left" gave you 6, and why that was wrong...)

Michael Weiss said...

Oops, please delete that last paragraph -- it was left over from a first draft and didn't get edited out. (Although it allows you to see my tracks: My first time through I focused on filling the offices, not the cubicles, which turns out to be a more complicated way of thinking about it. In particular I was mistaken in saying the overcount was a factor of 3*2, when it's actually just a factor of 2.)

Niels Henrik Abel said...

Catherine,

The basic counting rule comes into play this way:

Sometimes, instead of thinking "slots to fill," it's easier to think of "tasks to complete." In particular, that's how we want to think of this problem you've stated:

Task 1: Choose the two men (order unimportant)
Task 2: Choose the woman (order unimportant)

Basic counting rule:
(# of ways of performing task 1) X (# of ways of performing task 2)

Thus, we have (# of ways to choose 2 of 3 men) X (# of ways to choose 1 of 2 women), which is (using notation) C(3,2) X C(2,1) = 3 X 2 = 6

This takes care of the numerator of your probability fraction. The denominator, of course, is the total number of ways to choose 3 out of the 5 for the offices, without regard to either order or gender. Thus, the denominator of the probability fraction is C(5,3) = 5!/(3! X 2!) = 10.

Hence: probability of stated problem is 6/10 = 3/5.

Hope I've got that right ;) I will readily admit that counting problems can seem tricky if you're not accustomed to doing them. Every semester when I come to this section, I have to re-read the counting stuff to make sure that I remember enough of the details to be able to do these problems more or less off the cuff.

Anonymous said...

Regarding resources on this topic, I am finding Art of Problem Solving's book "Competition Math for Middle School" to be very helpful. It is straightforward, has many worked examples, and the problem sets have solutions explaining how to answer each problem.

-Karen

lgm said...

http://www.coolmath.com/algebra/20-combinatorics/index.html

also a straightforward free concept explanation

Anonymous said...

My explanation would be exactly the same as NHA's. Since the question is about the office, we can safely ignore the cubicle.

A mathematically equivalent question would be ....

"3 members of a table tennis team have to be chosen from 3 boys and 2 girls. What is the probability of the team having exactly 1 girl?"

(3C2 * 2C1) / 5C3 = 6/10 = 3/5

------------------------

After having said that, I also like to check the same answer with basic probability .... slot filling, if you will. If I can't do a question both ways and get the same answer ..... :-(

1st slot = boy = 3/5
2nd slot = boy = 2/4
3rd slot = girl = 2/3

There are three possible slots that the girl could be chosen for so ..

(3/5 * 2/4 * 2/3) * 3 = 3/5

Richard I