2.7 Chapter Summary

Investigate!

Suppose you have a huge box of animal crackers containing plenty of each of 10 different animals. For the counting questions below, carefully examine their similarities and differences, and then give an answer. The answers are all one of the following:

P open parenthesis 10, 6 close parenthesis

(\binom{10}{6})

10 to the power 6

(\binom{15}{9})

How many animal parades containing 6 crackers can you line up?
How many animal parades of 6 crackers can you line up so that the animals appear in alphabetical order?
How many ways could you line up 6 different animals in alphabetical order?
How many ways could you line up 6 different animals if they can come in any order?
How many ways could you give 6 children one animal cracker each?
How many ways could you give 6 children one animal cracker each so that no two kids get the same animal?
How many ways could you give out 6 giraffes to 10 kids?
Write a question about giving animal crackers to kids that has the answer $(\binom{10}{6})$ .

With all the different counting techniques we have mastered in this last chapter, it might be difficult to know when to apply which technique. Indeed, it is very easy to get mixed up and use the wrong counting method for a given problem. You get better with practice. As you practice you start to notice some trends that can help you distinguish between types of counting problems. Here are some suggestions that you might find helpful when deciding how to tackle a counting problem and checking whether your solution is correct.

Remember that you are counting the number of items in some list of outcomes. Write down part of this list. Write down an element in the middle of the list – how are you deciding whether your element really is in the list. Could you get this element more than once using your proposed answer?
If generating an element on the list involves selecting something (for example, picking a letter or picking a position to put a letter, etc), can the things you select be repeated? Remember, permutations and combinations select objects from a set without repeats.
Does order matter? Be careful here and be sure you know what your answer really means. We usually say that order matters when you get different outcomes when the same objects are selected in different orders. Combinations and “Stars & Bars” are used when order does not matter.
There are four possibilities when it comes to order and repeats. If order matters and repeats are allowed, the answer will look like $n^{k}$ . If order matters and repeats are not allowed, we have $P (n, k)$ . If order doesn't matter and repeats are allowed, use stars and bars. If order doesn't matter and repeats are not allowed, use $(\binom{n}{k})$ . But be careful: this only applies when you are selecting things, and you should make sure you know exactly what you are selecting before determining which case you are in.
Think about how you would represent your counting problem in terms of sets or functions. We know how to count different sorts of sets and different types of functions.
As we saw with combinatorial proofs, you can often solve a counting problem in more than one way. Do that, and compare your numerical answers. If they don't match, something is amiss.

While we have covered many counting techniques, we have really only scratched the surface of the large subject of enumerative combinatorics. There are mathematicians doing original research in this area even as you read this. Counting can be really hard.

In the next chapter, we will approach counting questions from a very different direction, and in doing so, answer infinitely many counting questions at the same time. We will create sequences of answers to related questions.

Chapter Review

For each of the following counting problems, say whether the answer is $(\binom{10}{4})$ , $P open parenthesis 10, 4 close parenthesis$ , or neither. If you answer is “neither,” say what the answer should be instead.

How many shortest lattice paths are there from $open parenthesis 0, 0 close parenthesis$ to $open parenthesis 10, 4 close parenthesis$ ?
If you have 10 bow ties, and you want to select 4 of them for next week, how many choices do you have?
Suppose you have 10 bow ties and you will wear a different one on each of the next 4 days. How many choices do you have?
If you want to wear 4 of your 10 bow ties next week (Monday through Sunday), how many ways can this be accomplished?
Out of a group of 10 classmates, how many ways can you rank your top 4 friends?
If 10 students come to their professor's office but only 4 can fit at a time, how different combinations of 4 students can see the prof first?
How many 4 letter words can be made from the first 10 letters of the alphabet?
How many ways can you make the word “cake” from the first 10 letters of the alphabet?
How many ways are there to distribute 10 identical apples among 4 children?
If you have 10 kids (and live in a shoe) and 4 types of cereal, how many ways can your kids eat breakfast?
How many ways can you arrange exactly 4 ones in a string of 10 binary digits?
You want to select 4 distinct, single-digit numbers as your lotto picks. How many choices do you have?
10 kids want ice-cream. You have 4 varieties. How many ways are there to give the kids as much ice-cream as they want?
How many 1-1 functions are there from $open brace 1, 2, and so on, 10 close brace$ to ${a, b, c, d}$ ?
How many surjective functions are there from $open brace 1, 2, and so on, 10 close brace$ to ${a, b, c, d}$ ?
Each of your 10 bow ties match 4 pairs of suspenders. How many outfits can you make?
After the party, the 10 kids each choose one of 4 party-favors. How many outcomes?
How many 6-elements subsets are there of the set $open brace 1, 2, and so on, 10 close brace$
How many ways can you split up 11 kids into 5 named teams?
How many solutions are there to $x_{1} + x_{2} + \dots + x_{5} = 6$ where each $x_{i}$ is a non-negative integer?
Your band goes on tour. There are 10 cities within driving distance, but only enough time to play 4 of them. How many choices do you have for the cities on your tour?
In how many different ways can you play the 4 cities you choose?
Out of the 10 breakfast cereals available, you want to have 4 bowls. How many ways can you do this?
There are 10 types of cookies available. You want to make a 4 cookie stack. How many different stacks can you make?
From your home at (0,0) you want to go to either the donut shop at (5,4) or the one at (3,6). How many paths could you take?
How many 10-digit numbers do not contain a sub-string of 4 repeated digits?

Neither. $(\binom{14}{4})$ paths.
$(\binom{10}{4})$ bow ties.
$P open parenthesis 10, 4 close parenthesis$ , since order is important.
Neither. Assuming you will wear each of the 4 ties on just 4 of the 7 days, without repeats: $(\binom{10}{4}) P (7, 4)$ .
$P open parenthesis 10, 4 close parenthesis$ .
$(\binom{10}{4})$ .
Neither. Since you could repeat letters: $10 to the power 4$ . If no repeats are allowed, it would be $P open parenthesis 10, 4 close parenthesis$ .
Neither. Actually, “k” is the 11th letter of the alphabet, so the answer is 0. If “k” was among the first 10 letters, there would only be 1 way - write it down.
Neither. Either $(\binom{9}{3})$ (if every kid gets an apple) or $(\binom{13}{3})$ (if appleless kids are allowed).
Neither. Note that this could not be $(\binom{10}{4})$ since the 10 things and 4 things are from different groups. $4 to the power 10$ , assuming each kid eats one type of cerial.
$(\binom{10}{4})$ - don't be fooled by the “arrange” in there - you are picking 4 out of 10 spots to put the 1's.
$(\binom{10}{4})$ (assuming order is irrelevant).
Neither. $16 to the power 10$ (each kid chooses yes or no to 4 varieties).
Neither. 0.
Neither. $4^{10} - [(\binom{4}{1}) 3^{10} - (\binom{4}{2}) 2^{10} + (\binom{4}{3}) 1^{10}]$ .
Neither. $10 times 4$ .
Neither. $4 to the power 10$ .
$(\binom{10}{4})$ (which is the same as $(\binom{10}{6})$ ).
Neither. If all the kids were identical, and you wanted no empty teams, it would be $(\binom{10}{4})$ . Instead, this will be the same as the number of surjective functions from a set of size 11 to a set of size 5.
$(\binom{10}{4})$ .
$(\binom{10}{4})$ .
Neither. $4 !$ .
Neither. It's $(\binom{10}{4})$ if you won't repeat any choices. If repetition is allowed, then this becomes $x_{1} + x_{2} + \dots + x_{10} = 4$ , which has $(\binom{13}{9})$ solutions in non-negative integers.
Neither. Since repetition of cookie type is allowed, the answer is $10 to the power 4$ . Without repetition, you would have $P open parenthesis 10, 4 close parenthesis$ .
$(\binom{10}{4})$ since that is equal to $(\binom{9}{4}) + (\binom{9}{3})$ .
Neither. It will be a complicated (possibly PIE) counting problem.

Give a counting question where the answer is $8 \cdot 3 \cdot 3 \cdot 5$ . Give another question where the answer is $8 + 3 + 3 + 5$ .

You own 8 purple bow ties, 3 red bow ties, 3 blue bow ties and 5 green bow ties. How many ways can you select one of each color bow tie to take with you on a trip? $8 \cdot 3 \cdot 3 \cdot 5$ ways. How many choices do you have for a single bow tie to wear tomorrow? $8 + 3 + 3 + 5$ choices.

Explain using lattice paths why $\sum_{k = 0}^{n} (\binom{n}{k}) = 2^{n}$ .

$2 to the power n$ is the number of lattice paths which have length $n$ , since for each step you can go up or right. Such a path would end along the line $x + y = n$ . So you will end at $open parenthesis 0, n close parenthesis$ , or $open parenthesis 1, n minus 1 close parenthesis$ or $open parenthesis 2, n minus 2 close parenthesis$ or … or $open parenthesis n, 0 close parenthesis$ . Counting the paths to each of these points separately, give $(\binom{n}{0})$ , $(\binom{n}{1})$ , $(\binom{n}{2})$ , …, $(\binom{n}{n})$ (each time choosing which of the $n$ steps to be to the right). These two methods count the same quantity, so are equal.

For which of the parts of the previous problem (Exercise) does it make sense to interpret the counting question as counting some number of functions? Say what the domain and codomain should be, and whether you are counting all functions, injections, surjections, or something else.

You are giving your professor 4 types of cookies coming from 10 different types of cookies. This does not lend itself well to a function interpretation. We could say that the domain contains the 4 types you will give your professor and the codomain contains the 10 you can choose from, but then counting injections would be too much (it doesn't matter if you pick type 3 first and type 2 second, or the other way around, just that you pick those two types).
We want to consider injective functions from the set ${$ most, second most, second least, least $}$ to the set of 10 cookie types. We want injections because we cannot pick the same type of cookie to give most and least of (for example).
This is not a good problem to interpret as a function. The problem is that the domain would have to be the 12 cookies you bake, but these elements are indistinguishable (there is not a first cookie, second cookie, etc.).
The domain should be the 12 shapes, the codomain the 10 types of cookies. Since we can use the same type for different shapes, we are interested in counting all functions here.
Here we insist that each type of cookie be given at least once, so now we are asking for the number of surjections of those functions counted in the previous part.

Discrete Mathematics: An Open Introduction, 3rd edition, by Oscar Levin (discrete.openmathbooks.org), licensed under CC BY-SA 4.0; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.