Allright, here they are:

1) Formulate the addition rule and multiplication rule in your own words

**The addition rule: **

Suppose you have n_{A} choices for A and n_{B} choices for B. If none choices for A and B coincide, then number of ways to choose from A either B is n_{A}+n_{B} (hence the name "Addition rule").

** Multiplication rule **

Suppose you have n_{A} choices for A and n_{B} choices for B. The number of ways to choose A AND B is n_{A}n_{B} (hence the name "multiplication rule").

For short, you can memorize that "OR is PLUS and AND is TIMES".

2) What is the number of different passwords of length 5:

a) containing digits only

b) containing digits OR one of 10 characters , . / ? ' " ! * ( )

c) how much the answer in b) is bigger than the answer in a) (you have to give an explicit answer here)

**The solution: **

a) There are 10 digits, and we have to choose 5 times out of 10. We apply multiplication rule. We have to choose the first digit AND the second digit - by multiplication rule, we can do it in 10*10=100 ways. And, we also have to choose the third digit, this come up to 100*10=1000 ways. And so on, in total we have 10*10*10*10*10=10^{5} ways.

b) For every position we have 10 digits and 10 characters, in totoal 20 choices/ By the very same logic as in a), the answer for b) is 205

c) We are asked to calculate \( 20^5/10^5 \) . Doesn't that go over 40 minutes of the the time alotted? No. The smarter you get, the less you have to count. Look: \[ 20^5=(2*10)^5=2^5*10^5 \] , hence

\[ \frac{20^5}{10^5}=\frac{2^5*10^5}{10^5}=2^5 \]

And 2^{5} is not too complicated to calculate - it's 32.

3) Joe wants to throw a party. He has to choose a stuffing for quesadillas out of 3 different ones and a drink out of 5 options (but keep in mind that all of them are non-alcoholic!). In how many ways can he make a choice (please give explicit number)?

**The solution: **

This is canonical case for multiplication rule: 3 different quesadilla stuffings and 5 different non-alcoholic drinks. So, the answer is 15.

4*) Mary is organizing her vacations. She will go to Florida, New York, Califormia, Mexico and Samara, Russia once. In how many ways can she order her destinations (please give explicit number)?

**The solution: **

This is exactly number-of-permutations problem. We have 4 different objects, and we have to put them into 4 different places (one of the places has to go first, another one - second, etc.). The answer is 4!=24, but let's remind ourselves how to derive it.

So, Mary can choose where to go first by 4 ways. She has to also choose the second place to go. This is "and" situation, so we use multiplication rule. She can choose the second place to go by 3 ways (because she's already chosen the first one). For the third place to go, there are only 2 options left, and for the last place to go, she has only one choice. So, the answer is 4*3*2*1=4!=24.

5) Please choose quadratic functions from the following:

a) \( x^2+2x+1 \)

b) \( x^3+2x+1 \)

c) \( (x-1)^2+1 \)

d) \( -2(x+0.5)^2-0.1 \)

e) \( (x+10)^3+1 \)

f) \( (x-1)(x-3) \)

g) \( x^2+x \)

h) \( -2x^2+1 \)

j) \( x+153.5 \)

Out of those which are quadatic functions, choose the ones whose branches look up.

**The solution: **

The quadratic function is a polynomial of a degree 2, i. e. it has (or can be made to have) a form \( ax^2+bx+c \) with nonzero \(a\) . Out of all options, only b), e) and j) do mot comply with the definition. In b), we have x^{3}, that means that it's cubic function. In j), we don't have quadratic term \(ax^2\) , so it's linear function.

The general rule for \( ax^2+bx+c \) function is: if \( a > 0 \) then the branches go up, and if \( a < 0 \) the branches go down. So, for a), c), f), g) the branches go up, and for d), h) the branches go down. In order to see that, in c), d), f) we have to expand the brackets.