# Harrison Chapman

## Math 301: Homework 1

Due: Friday, September 6, 2019

While you don’t have to write formal proofs for many solutions, if you’re only writing symbols, you’re not explaining things. Make sentences by adding in a few English words to explain your reasoning to me.

1. How many ways are there to pair up 5 students into two groups of two (leaving one odd-student-out)?

There are $$\binom 5 2$$ ways to form a first pair, then $$\binom 3 2$$ to form the second pair. There are 2 pairs, and their order doesn’t matter, so the answer is:

$\binom 52 \binom 32 \frac{1}{2!}$

2. How many subsets does that set $A = \{1,2,3,4,5\}$ have? How many of the subsets have cardinality 3?

There are $$2^{\vert A \vert} = 2^{5}$$ subsets of $$A$$, of which $$\binom 5 3$$ have cardinality 3.

3. Let $A$ and $B$ be sets. If $A \cap B = \emptyset$, what is $A \setminus B$? $A \Delta B$?

As they are disjoint, $$A \setminus B = A$$ and $$A \Delta B = A \cup B$$.

4. Explain why $\vert A \cap B \vert + \vert A \cup B \vert = \vert A \vert + \vert B \vert$. A picture will help.

We’ll use the fact that if $$S, T$$ are disjoint then $$\vert S \cup T \vert = \vert S \vert + \vert T \vert$$.

First notice that $$(A \setminus B)$$ and $$(A \cap B)$$ are disjoint and $$(A \setminus B) \cup (A \cap B) = A$$. So $$\vert A \vert = \vert A \setminus B \vert + \vert A \cap B \vert$$. For similar reasons, $$\vert B \vert = \vert B \setminus A \vert + \vert A \cap B \vert$$.

Now notice that $$A \cup B = (A \setminus B) \cup (A \cap B) \cup (B \setminus A)$$, where the three sets on the right hand side are all disjoint. So:

So $$\vert A \cup B \vert = \vert A \vert + \vert B \vert - \vert A \cap B \vert$$, but this is equivalent to what we were trying to prove.

5. If one were to roll 4 identical 3-sided dice (ignoring for a moment how a 3-sided die would look!), how many outcomes are possible?

To any dice roll, we could sort the dice in a row so that the lower-valued dice are to the left of the higher-valued dice. We could put toothpicks between dice of a value difference by 1 (and so 2 if value difference of 2, and so on).

There are $$4 + (3-1)$$ slots to be filled with dice and toothpicks, and we place $$(3-1)$$ toothpicks because the dice take on 3 different values separated by 1. So the number of outcomes is:

$\binom {4 + (3-1)}{3-1}$