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Math Notes

Counting Methods, With Repetition, Order Doesn't Matter

Introduction

Building from the Fundamental Counting Principle described in Counting Methods I, we want to now address the situation in which repetition is allowed, but order does not matter.   This is the case they usually don't teach you in school.  You may choose to move on to Counting Methods III without loss of context.  Let's consider the case of the Pick3 Lottery as our example.

One way of winning the Pick3 Lottery is to have all the numbers that are chosen, but in any order.  Each digit can be any number at all, but for this particular method of winning, the order these numbers appear in is irrelevant.  So let's pick a particular set of numbers and see what happens.  Let's suppose that all the digits are different, like 123.  Let's list all the possible combinations of these three digits.

123       132       213       231        312         321

Each one of these numbers represents a single winning number combination.  There are six of them.

The three digit case has six numbers that represent a single winner or 1*2*3.  The two digit case (described in example A) has two numbers that represent a single number, or 1*2.  The four-digit case (described in example B) has 24 numbers that represent a single winner, this is 1*2*3*4.  This pattern of numbers is referred to as factorials.  Factorials can be written shorthand as 2! 3! or 4!.  The ! tells us we are going to start with the largest number listed, the one next to the !, and then multiply by all the numbers less than that until we get to one.  One note, 0! is defined to be 1 also.

To finish these problems, we need to know how many ways can we win by matching the numbers but not the order of the winning number.  What we have to do is divide our problem up into three parts and use the pattern we just observed.  The three parts of our problem are: no repeated numbers, one repeated numbers, all repeated numbers.

For the case with no repeated numbers, there are 10 ways to choose the first digit, 9 ways to choose the second, and 8 ways to choose the third (by the Fundamental Counting Principle) or 720.  But this number assumes order matters, so we are going to divide as we saw before by 3! because there are six numbers that match each winning number.  Leaving us with 120 winners where all three numbers are different.

The case where all the numbers are the same, say 111, can only match if all the numbers are alike, and the match is spot on.  There are 10 possibilities for the first number, and one choice for the remaining two digits.  So there are 10 winners.

For the case of one repeated number, let's take a single example again, like 122.  How many combinations are there?

122     212    221

There are just three numbers that belong to a single winner.  How many of these winners are there?  We can shortcut the counting process by adding up our other two cases and subtracting them from the total number of three digit numbers of any kind.  There are 1000 of those (10*10*10) and we've used up 720 and another 10, so 1000 - 720 - 10 = 270.  270 is the total numbers of this type, but 270/3 = 90 is the number of tickets we'd have to buy to cover them.

To find the total number of winners, we add them all up.  120 + 90 + 10 = 220.

We'll do the examples of two and four digits in the Worked Examples section.  Five digits is in the Challenge Problems section.  You can follow the example of how I built the four digit case off of the three digit case to build the five digit case off of the four digit case.

Worked Examples

A. Let's consider the example of an imaginary Pick2 game which we can win in the same way as above.  Consider the same problem, but with only two digits, like 12.  All the possible combinations are:

12    21

Each of these numbers represents a single number combination.  There are two of them.

Each two digit number with different digits and it's counterpart in the reverse order represents a single winner.  So, 12 and 21 are the same winner.  Using the Fundamental Counting Principle, we know that there are 10 values for the first digit and 9 for the second.  We need to divide by two to get the number of winners of this type, so 10*9/2 = 90/2 = 45.  There are also ten cases where the digits match, 00, 11, 22, etc.  So 10 + 45 = 55 possible winners.

B. What if we considered the Pick4 Lottery.  One way of winning is the same as with the Pick3, match all the numbers drawn, in any order.  So consider a winning number like 1234.  All the possible numbers that would win under this method are:

1234    1243    1324    1342   1423    1432   2134    2143   2314    2341    2413   2431    3124    3142    3214    3241     3412     3421    4123    4132    4213     4231    4312    4321

Each one of these represents a single winning number combination.  There are 24 of them.

What we established above was the case when none of the digits matched.  There are 24 matches for each winner;  to find the number of winners of this type, there are 10 values for the first digit, 9 values for the second, 8 for the third, and 7 for the fourth.  We get 10*9*8*7/24 = 5040/24 = 210.

What if two of the digits match?  Well, like the case with three digits, let's assume that it's the last digit we're matching, for instance, 1233.  What are all the possible combinations of these?

1233    1323     1332    2133   2313    2331   3123    3132    3213    3231    3312    3321

First we find the number of ways to generate a number like the one above, where we have three non-matching digits, and then one matching the last digit and then we divide by 12 because there are 12 different orders of these numbers that all represent a single winner.  Notice that's 12 = 3*4.  Now we have to find out how many numbers satisfy this condition.  We can use what we know of the Pick3 game and a tree diagram.

There are two ways to get numbers where just a single pair of the digits match.  Either we start with a three digit number with no match and add a digit that's already there, or we start with a three-digit number where there is a single matching pair and we add something which doesn't appear in the number.  We saw from the Pick3 game that there are 720 numbers with no match, and to match one of them, we have three ways to do that, so we get 720*3 = 2160.  For the second case, there are 270 numbers with a pair of matching digits.  So that we don't produce another pair or a threesome have to use something we haven't used yet, so there are 8 possibilities.  That's 270*8 = 2160.  Add these up and we get 4320.  If we now divide this by 12 we get the number for this case which is 4320/12 = 360.

We also have to consider what happens with numbers like 1212 and 1222:

1122    1212    1221    2112     2211   2121

1222    2122     2212    2221

The only way to get a double matching pair is to start with a three digit number which already has a match, of which there are 270, and we need to add the one number which isn't part of a pair yet.  We can do that only one way, so 270*1 = 270 numbers with two pairs of matching digits.  Since there are six different ways we can order this two pairs, we divide by six and get 270/6 = 45.

For the three digit match, there are two ways to produce these numbers. We can start with a three-digit number that already has three-of-a-kind, and add a different digit.  There are three-digit numbers of this kind, and we add a digit which isn't the one already used so we have nine choices for this new digit.  10*9 = 90.  Also, we can start out with a number with a pair of matching digits and add another one of those digits.  There are 270 three-digit numbers with a matching pair, and only one option to make this three-of-a-kind, so 270*1 = 270.  If we add these up, we get 270 + 90 = 360.  We now divide by 4 since there are four ways to organize these digits, which gives us 360/4 = 90.

And the last case we have to consider is with the case where all the digits are the same, like 1111.  There are 10 of these.

Now add up all the cases.  In summary: No Match = 210; Pair of matching digits = 360; Two pairs of matching digits = 45; Set of three matching digits = 90; All digits matching = 10.  So 210 + 360 + 45 + 90 + 10 = 715.

C. Suppose now we're playing a funny new version of Pick3.  Let's suppose it's set up in such a way to produce only natural three digit numbers, i.e. no leading zeros.  How many tickets do we need to buy now to guarantee a match if order doesn't matter?

The first thing we have to figure out is the total number of possible winning numbers we are working with if order mattered.  We can do this through the fundamental counting principle.  We have only 1-9 as options for the first digit, but still 0-10 for the other two.  So that's 9*10*10 = 900 instead of 1000, which we had for the usual Pick3 game.

Then we have to break our problem up into several smaller problems, sets of numbers that will behave similarly.  Previously, we had three sets: no repetitions, one repetition, and all digits identical.  Some of our numbers will behave like these if we take zero out of the mix entirely.  The rest of the numbers will have to deal with separately--whenever we have special restrictions we will end up with these special extra cases.  With no zero, our possible winners (when order matters) is now 9*9*9 = 729.  Our three sets now have a maximum size of: no repeats -- 9*8*7 = 504; all the same = 9; one repetition -- 729 - 504 -9 = 216.

To determine the total number of tickets we'd need to buy to cover these subsets, we would continue as we had in the regular Pick3 game.  For the set with no repeated digits, like 123, we can arrange these in six ways

123      132      213       231       312       321

So we divide 504 (the total number of this type) by six, 504/6 = 84.

For the case with all the digits repeated, we have to buy one for each, that's still 9.  For the case with one repeated digit, like 122

122          212           221

There are three numbers using these digits so we'll divide 216 by three.  216/3 = 72.

We still haven't touched the numbers containing a zero yet, and there are 900 - 729 = 171 of those.  Consider a number like 102.  How many numbers are like this?  9*1*8 = 72 (we're treating the case with two zeros as a separate case, that's why there are only eight possibilities for the last digit).  We add to these the numbers that have only one zero, but at the end, like 120.  Like the previous calculation, there are 9*8*1 = 72 of them, so the total number of winners with one zero and no repetition is 72 + 72 = 144.  How many ways can we organize these?  Consider a number like 301.

103      130       301       310       013       031

We have to toss out the cases with leading zeros, giving us only four ways to organize these digits.  So if we do the division, 144/4 = 36.

Now about the other cases with repeated digits?  Non-zero repeated digits like 110, and 101.  We get 9*1*1 = 9 and 9*1*1 = 9 for each type or 9 + 9 = 18.  We can arrange them as

101          110            011

Throw out the last since it starts with zero, so now we divide 18/2 = 9.

The last case is the case with double zeros, like 100.  There are 9*1*1 = 9 of them, and can be arranged as

100             010             001

We throw out the cases that start with zero, so we keep all 9.  That's all the cases.  Now add them up.  84 + 9 + 72 + 36 + 9 + 9 = 219.

[Something to notice about this problem is that even though we dropped one hundred possible winning numbers, we had to buy nearly as many tickets, only 219 instead of 220.  The only ticket we didn't have to buy was 000.]

Problem Solving Tips

Solving problems of this type are generally particularly complex.  This is one of the reasons why it is so infrequently addressed in basic courses.  You will see The Fundamental Counting Principle, Permutations and Combinations frequently, but problems of this sort are not that common.  As you can see, once you are familiar with combinations, there is a certain similarity to these problems, but less straightforward.  To solve problems of this type you are going to have to look for a couple of things:

• Just as with simpler problems, read the problem all the way through and then find what it is exactly that you're looking for.
• As part of figuring out what you're looking you, are are going to need to determine if this problem is a one-step problem, or a multi-step problem.  If it's a multi-step problem, you'll need to determine how many different parts there are.  Ask yourself if any parts of the problem behave differently than other parts; for example, numbers with no matching digits, versus numbers with matching digits.
• You may find it useful to build up cases using a tree diagram.
• For each part of the problem you need to determine two things: the number of total ways you can satisfy the condition, if order mattered, and secondly, the total number of ways you can organize the digits or letters (or whatever determines the order).  Sometimes you will be able to do this by formula, sometimes you will need to do this by listing the possibilities but you should try to be methodical about it.  If one method doesn't work, try another.  Once you find these two values, you will need to divide the larger by the smaller for each case.
• Make sure you use a methodological approach to listing all the ways the digits can be arranged.  Try breaking the problem up into small units, say the first two or three digits, and the second two or three digits, etc.  Use this methodology to list out the cases.  When repetition occurs, the Fundamental Counting Principle doesn't work as straightforwardly.
• If you're using a TI-83 Calculator, you can find the factorial key by selecting the Math menu (it has it's own button), scrolling over to the PRB menu.  It's the first command.  Remember to type the number you want, and then hit factorial, then enter.
• Don't forget to add up all the cases to get your final answer.

Additional Problems with Answers

1. Suppose you want to determine the number of ways you can win a big state lottery.  You know there is only one winner if you want to match the numbers in the correct order, so you decide to just try to get the numbers right, but in any order.  You need to choose six numbers from 42.  There is no possibility of repeating numbers.

2. You would like to get four of a kind in a poker game.  How many different four-of-a-kind hands are there?

3. I'm flipping a coin.  How many ways are there to flip the coin five times and get exactly three heads?

Challenge Problems:

4. The Anywhere State Lottery has a five-digit kicker number on their regular lottery tickets that you can purchase for an extra buck.  It's like a Pick5.  If you aren't concerned about matching the order, how many tickets would you need to buy to guarantee a win?

Answers

1. 5,245,786
2. 5760 [Hint: separate this problem into three parts: the part with the Aces, the spare card, and the order that the spare card can be drawn in between the aces.]
3. 10
4. 3754 [Hint: there are seven different cases you have to consider, no matching digits, 2 matching digits, 2 pairs and a spare, one pair/one threesome, three-of-a-kind, four-of-a-kind, and all five matching.  You can use a tree diagram based on the Pick4 above to get 30240, 54720, 6480, 900, 7200, 450 and 10 respectively if order matters.  If you if you check the number of possible orders, you find that there are 120 matching orders, 20, 30, 10, 20, 5 and 1 respectively for each of the cases.  Now divide and add.]

Links to Outside Sources

7.6 - Counting Principles

Links to Supporting Topics

Counting Methods I -- With Repetition, Order Matters: Fundamental Counting Principle
Counting Methods III -- No Repetition, Order Matters: Permutations
Counting Methods IV -- No Repetition, Order Doesn't Matter: Combinations

History

If you know of any links the discuss the history, origin or development of this method, or any other short-cuts, please email me at the address below.