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Midterm

Paper 2017

Quiz-1

Two "words" are considered equivalent if one is a rearrangement of the other. (Cat, act and cta are considered same for example). How many three-letter words that are not equivalent can you make?

We can choose 3 items from 26 letters and two spaces, so it's 26+2\choose 3, or 28\choose 3.

Quiz-2

Provide a 2 column proof: if A \subseteq B, then \complement_U{A} \cup B = U

To show \complement_U{A} \cup B = U, we need to prove \complement_U{A} \cup B \subseteq U and U \subseteq \complement_U{A}\cup B

  1. \complement_U{A} \cup B \subseteq U is obvious by the definition of U.
  2. To show U \subseteq \complement_U{A}\cup B:
\begin{align} & \text{Let } x\in U \Rightarrow x\in A \text{ or } x\in \complement_U{A}. &&\text{Definition of }U\\ & \text{If } x\in A \Rightarrow x\in B \Rightarrow x\in \complement_U{A}\cup B &&A \subseteq B\\ & \text{If } x\in \complement_U{A} \Rightarrow x\in \complement_U{A}\cup B &&\text{Definition of operator }\cup\\ \end{align}

Quiz-3

Use a 2-column and rules of inference with quantifiers to verify the validity of the following:
1. All wizards at school must know how to make potions.
2. All wizards who do magic outside of school must know how to cast spells.
3. Harry is a wizard at school, but he doesn't know how to cast spells.
4. Hermione knows How to cast spells but doesn't know hot ot make potions.
Therefore Harry does not do magic out side of school and Hermione isn't a wizard at school.

p(x): x is a wizard at school.
q(x): x knows how to make potions.
r(x): x do magic outside of school.
s(x): x knows how to cast spells.

\begin{align} (1) \text{ }& \forall{x}[p(x) \rightarrow q(x)] && \text{given} \\ (2) \text{ }& \forall{x}[(p(x)\land{r(x)}) \rightarrow s(x)] && \text{given} \\ (3) \text{ }& p(Harry) && \text{given} \\ (4) \text{ }& \lnot{s(Harry)} && \text{given} \\ (5) \text{ }& (p(Harry)\land{r(Harry)}) \rightarrow s(Harry) && \text{(2) Universal Spec} \\ (6) \text{ }& \lnot (p(Harry)\land{r(Harry)}) && \text{(4)(5) M. Tollen's} \\ (7) \text{ }& \lnot p(Harry) \lor \lnot{r(Harry)} && \text{(6) DMorgan} \\ (8) \text{ }& \lnot{r(Harry)} && \text{(3)(7) Disjunctive Syllogism} \\ (9) \text{ }& \lnot q(Hermione) && \text{given} \\ (10)\text{ }& p(Hermione) \rightarrow q(Hermione) && \text{(1) Universal Spec} \\ (11)\text{ }& \lnot p(Hermione) && \text{(9)(10) M. Tollen's} \\ (12)\text{ }& \lnot{r(Harry)} \land \lnot p(Hermione) && \text{(8)(11) Conjunction} \\ \end{align}

Quiz-4

How many ways are there to place 10 books on 5 bookshelves, if the order of books relevant and (a) shelf can be empty or (b) shelves can be left empty.

Because the order of books relevant, we arrange books first, which is P(10, 5) = 5.

If shelves can be empty, then we are assigning these 4 bars in 5 categories to 10+5-1 places, which is 14\choose 4, so number of ways are: $$ P(10, 5) \times {14\choose 4} = \dfrac{10!14!}{5!10!4!} = = \dfrac{14!}{5!4!} $$

If shelves cannot be empty, then we are assigning these 4 bars in 5 categories to 10-1 places, which is 9\choose 4, so number of ways are: $$ P(10, 5) \times {9\choose 4} = \dfrac{10!9!}{5!5!4!} $$

Quiz-5

How may permutations of the 26 different letters of the alphabet contain the pattern "DOG", the pattern "CAT", or the "MICE" pattern?

Individual permutations:

  • "DOG": D = P(24, 24) = 24!
  • "CAT": C = P(24, 24) = 24!
  • "MICE":M = P(23, 23) = 23!.

The overall permutations is C\cup D\cup M, mind that elements included "DOG" and "CAT", "DOG" and "MICE" are partially overlapping. So in order to eliminate duplications, I use the Venn diagram helping me understand.

Credit to MyDraw

Venn diagram obviously shows that: $$ C\cup D\cup M = C + D + M - C\cap D - C\cap M - D\cap M + C\cap D\cap M $$

Since:

\begin{align} C\cap D &= P(26-6+2, 26-6+2) = P(22, 22) = 22!\\ C\cap M &= 0\\ D\cap M &= P(26-7+2, 26-7+2) = P(21, 21) = 21!\\ C\cap D\cap M &= 0 \end{align}

So:

\begin{align} C\cup D\cup M &= C + D + M - C\cap D - C\cap M - D\cap M + C\cap D\cap M\\ &= 24! + 24! + 22! - 22! \end{align}

Quiz-6

Prove S(n): \sum_{i=1}^n f_i^2 = f_nf_{n+1} for all n, where f_0 = 0, f_1 = f_2 = 1, f_{n+2} = f_n + f_{n+1}.

Quiz-7

Let a, b, c, d be positive integers. Prove [(a|b)\land(c|d)\Rightarrow ac|bd].

\begin{align} & (1) && a|b \Rightarrow b = an, n\in \mathbb{Z} && \text{ Definition}\\ & (2) && c|d \Rightarrow d = cm, m\in \mathbb{Z} && \text{ Definition}\\ & (3) && bd = (an)(cm) = (ac)(mn), mn\in \mathbb{Z} && \text{ (1)(2)}\\ & (4) && ac|bd && \text{ (3) definition}\\ \end{align}

Quiz-8

Negate and simplify: \forall x\exists y[(p(x, y)\land q(x, y))\rightarrow r(x, y)], use 2 column proof format.

\begin{align} \lnot[\forall x\exists y[( p(x, y) \land q(x, y)) \rightarrow r(x, y)]&] &&\text{ Negate}\\ \exists x\lnot[\exists y[( p(x, y) \land q(x, y)) \rightarrow r(x, y)]&] &&\text{ Negation of }\forall\\ \exists x\forall y\lnot[( p(x, y) \land q(x, y)) \rightarrow r(x, y)]& &&\text{ Negation of }\exists\\ \exists x\forall y\lnot[\lnot(p(x, y) \land q(x, y)) \lor r(x, y)]& &&\text{ Logically equivalent}\\ \exists x\forall y[\lnot\lnot(p(x, y) \land q(x, y)) \land \lnot r(x, y)]& &&\text{ d'Morgan}\\ \exists x\forall y[( p(x, y) \land q(x, y)) \land \lnot r(x, y)]& &&\text{ Double negation}\\ \end{align}

Quiz-9

n/a

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