The **Rule of Sum (Addition Principle)** and **Rule of Product (Multiplication Principle)** are principles of counting that are used to build up the theory of enumerative combinatorics.

Rule of Sum / Addition principle:If there are $n$ ways of doings something, and $m$ ways of doing another thing, both of which cannot be done at the same time, then there are $n+m$ ways to choose one of these actions.

Rule of Product / Multiplication principle:If there are $n$ ways of doing something, and $m$ ways of doing another thing after that, then there are $n\times m$ ways to perform both of these actions.

## 1. Calvin wants to go to Milwaukee. He can choose from $3$ bus services or $2$ train services to head from home to downtown Chicago. From there, he can choose from 2 bus services or 3 train services to head to Milwaukee. How many ways are there for him to get to Milwaukee?

Solution: Since Calvin can either take a bus or a train downtown , he has $3+2 =5$ ways to head downtown (Rule of Sum). After which, he can either take a bus or a train to Milwaukee, hence he has another $2+3=5$ ways to head to Milwaukee (Rule of Sum). Thus in total, he has $5 \times 5 = 25$ ways to head from home to Milwaukee (Rule of Product).

## 2. Calvin wants to go to Milwaukee (see previous question). This time, he has to purchase a bus concession (which will only allow him to take buses), or a train concession (which will only allow him to take trains). If he only has money for $1$ of these concessions, how many ways are there for him to get to Milwaukee?

Solution: If Calvin purchases a bus concession, he has $3 \times 2=6$ ways to get to Milwaukee (Rule of Product). If Calvin purchases a train concession, he has $2\times3=6$ ways to get to Milwaukee (Rule of Product). Hence, he has $6+6=12$ ways to get to Milwaukee in total (Rule of Sum).

## 3. Six friends Andy, Bandy, Candy, Dandy, Endy and Fandy want to sit in a row at the cinema. If there are only six seats available, how many ways can we seat these friends?

Solution: For the first seat, we have a choice of any of the 6 friends. After seating the first person, for the second seat, we have a choice of any of the remaining 5 friends. After seating the second person, for the third seat, we have a choice of any of the remaining 4 friends. After seating the third person, for the fourth seat, we have a choice of any of the remaining 3 friends. After seating the fourth person, for the fifth seat, we have a choice of any of the remaining 2 friends. After seating the fifth person, for the sixth seat, we have a choice of only 1 of the remaining friends. Hence, by the Rule of Product, there are $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ ways to seat these 6 people. More generally, this problem is known as a Permutation. There are $n! = n \times (n-1) \times (n-2) \times \ldots \times 1$ ways to seat $n$ people in a row.

## 4. How many positive divisors does $2000 = 2^4 5^3$ have?

Solution: Any positive divisor of 2000 must have the form $2^a 5^b$, where $a$ and $b$ are integers satisfying $0 \leq a \leq 4, 0 \leq b \leq 3$. There are 5 possibilities for $a$ and 4 possibilities for $b$, hence there are $5 \times 4 = 20$ (Rule of Product) positive divisors of 2000 in all.

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TopNewestThe last two of them really helped thank you!

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