Counting Techniques and Some Other Math Team Strategies Susan Schwartz Wildstrom Walt Whitman High School Bethesda, MD NCTM National Meeting April 6, 2001.

Counting Techniques and Some Other Math Team Strategies

Susan Schwartz Wildstrom

Walt Whitman High School

Bethesda, MD

NCTM National Meeting

April 6, 2001

Orlando, FL

COUNTING

Review of familiar ideas

If we have n distinct objects and we want to put them in some sort of ordered set (arrangement) we use permutations, denoted or and equal to n nP ,n nP

! 1 2 3 ...2 1n n n n n

If we have n distinct objects and we want to order some subset (r) of them, we again want permutations, but this time we want to use just part of the set and denote this or which equals

Notice that you have r factors in this expansion.

n rP

,n rP

!1 2 ... 1

1 !

nn n n n r

n r

HARDER THINGS TO COUNT

Suppose that we still have n objects in our set but that some of them are indistinguishable from one another (an example of this would be the set of letters in the word “googol”). Suppose that we are interested in using all of the letters for each arrangement but we want to know how many distinguishably different arrangements there can be.

Here we want to use permutations again but we want to divide out those permutations which are not distinguishable from one another by virtue of their containing some of the repeated objects in different locations. Suppose there are p of one object, q of another, etc. Our calculation can be done as !

! !

n

p q

Suppose we have m identical objects to place in n (n<m) “boxes. If each box must contain at least one object, the task is like placing n-1 partitions in the m-1 spaces between the individual objects.

But, if it is acceptable for one or more of the boxes to be empty, the task is a bit more difficult. We can find the number of possibilities here by using the m objects and n additional “virtual” objects, ie, m+n objects. We then place n-1 partitions in the m+n-1 spaces between the objects, “remove” the one virtual object from each box and have the desired arrangement. Computationally then, this is

1

1

m

n

1

1

m n

n

Paths on a grid

Suppose we want to compute the number of paths there are from point A to point B travelling only along grid lines and through lattice points, each step moving us closer to B. And suppose, further that B is a point that is to the right and above A on a grid. (For simplicity, let’s place A at the origin (0,0).)

If B(m,n), then the task consists of a total of m+n steps each of which is either a move to the right or a move up. So, counting the number of paths that can be found consists of counting the number of ways that m moves to the right can be taken in a sequence of m+n total moves (not coincidentally, this is the same number that we get if we count the number of ways that n moves upward can be taken in m+n moves.)

So to find the number of possible paths, we need only compute the number of ways to choose m from m+n, ie,

Now, what if we want to insure that we pass through some specific point C (between A and B) on our way? Just find the product of the number of paths from A to C and the number of paths from C to B.

m n

m

DerangementsIf we have n numbered objects and n numbered positions for these objects, there is exactly one (and only one) way for each object to go into its correct position. There are a lot of ways for some of the objects to be correctly placed and some others to be out of place. But a more interesting question now is, how many ways are there for all of the objects to be in the wrong position?

This problem is much more difficult than it looks at first glance. There are two reasonable ways to analyze this situation--count them directly or count them by developing a recursive pattern.

Let’s do a few of the early derangements directly and learn a bit about inclusion-exclusion at the same time.

For n=2, we have 12 or 21. Only one of these is a derangement.

For n=3, we have:

123

132

213

231

312

321

Of these 231 and 312 are derangements.

n D(n)1 02 13 24567

Let’s list the permutations for 4:

1234 2134 3124 4123

1243 2143 3142 4132

1324 2314 3214 4213

1342 2341 3241 4231

1423 2413 3412 4312

1432 2431 3421 4321

How many of these are derangements? This is getting pretty hard to count by hand.

n D(n)1 02 13 24 9567

It’s time to organize our thoughts in a more efficient way. (Let’s look at n=4 again)

1. There are 4! = 24 orderings.

2. There are 6 with a 1 in position 1, 6 with a 2 in position 2, 6 with a 3 in position 3, and 6 with a 4 in position 4. Subtract these from the total number of orderings. (24-24)

3. But now we have subtracted too many since, for example, 1243 has both a 1 in position 1 and a 2 in position 2. So now we need to add back in the ones that have two numbers in the right positions because we subtracted them out twice.

There are 6 pairings of digits that could have both been in the correct place (12, 13, 14, 23, 24, 34) and each such pairing occurs in 2 numbers (eg, 1234 and 1243) so add back in

4. But now we’ve added some of them back in extra times, namely those with three numbers in the right positions (there are 4 such sets, each occurring 1 time). To undo this, we subtract out .

6 2

4 1

5. Then, lastly, add back in the one we have subtracted out too often, the one that has all four in the right position.

So what we have constructed now looks like this:

4! 4 6 6 2 4 1 1 9

Analyze 5 this way:

Total number of possible arrangements

- arrangements with 1 correct

+ arrangements with 2 correct

- arrangements with 3 correct

+ arrangements with 4 correct

- arrangements with 5 correct

5 5 5 5 55! 4! 3! 2! 1! 0!

1 2 3 4 5

120 120 60 20 5 1 44

Notice that the first factors of each term after the first are binomial coefficients for n, and the second factors of each are the permutations of the remaining quantity of numbers. Notice that we alternately subtract and add (inclusion-exclusion) to counteract the excesses of the previous counting operation.

You try 6 on your own and see whether you can come up with the right value, which is 265.

Notice how quickly the numbers get large. Also, notice that there doesn’t seem to be an obvious pattern. But we can find a recursive pattern. Do you see one?

n D(n)1 02 13 24 95 446 2657

1 1n

D n n D n

Finding the values of trig functions for angles of 15 and 75

Sometimes we need to use the trig function values for the semi-special angles of 15 and 75 (/12 and 5/12).

If you really know your trig formulas you can derive them using sum and difference or half-angle formulas.

Now we are going to look at a diagram that you can draw anytime you need them that will show them all to you at once.

Label all of the missing sides and angles of this rectangle based on what you know about the

45-45-90 and

30-60-90 right triangle patterns, and you get all the extra information about 15 and 75 as well.

45

45

30

601

45

45

45

45

45

45

30

60

15

75

90

90

90 90

1

23

62

22

22

62

6 22

6 22

And lastly, there is this old standby, (No calculators, of course.) a problem in which finding the final answer requires an intermediate evaluation step like simplifying

Whenever we see something like this in a math competition we are pretty sure that it will simplify, but how?

114 36 10

We can “experiment” but most of the time that will be too far-ranging, time-consuming and unproductive, so let’s develop a strategy that will solve the problem directly.

Suppose that

(And in this case, it’s since the value inside the surd is a difference.)

114 36 10 a b a b

2 2

114 36 10

114 36 10 2

a b

a ab b

Regardless of whether a and b are rational or irrational will be rational. So,

Solve the system: Use

2 2a b2 2 114

2 36 10 18 10

a b

ab ab

18 10a

b

2

2

4 2

4 2 2 2

2 2

3240114

3240 114

114 3240 60 54 0

60 or 54 and 54 or 60

bb

b b

b b b b

b a

Since

we want

So,

114 36 10 0a b a b

114 36 10 60 54 2 15 3 6