Math 10B Midterm 1 True/False

When a student is sick in the morning and might feel better later in the day, it is OK to send an email to the instructor to request to take a quiz in a later section because such a question is not addressed in the syllabus, make-up quizzes in other sections are allowed in this class, and the professor can quickly answer any number of such emails in time for the later quizzes.

False

k x nCn = n x n-1Ck-1 unless k > n.

False

Asking your GSI to drop more than 3 quiz scores is OK because someone can miss more than 3 sections for important reasons and in addition we would not want to keep low quiz scores due to poor quiz performance or lagging behind with the material.

False

It is always true that floor function of x is less than or equal to x which is less than or equal to the ceiling function of x, but equality of the two extreme terms of this inequality is never possible.

3. False - Even though the inequalities is TRUE for all real numbers, the latter part of the statement is FALSE. Equalities hold for all integers. (0 as a counterexample)

Any version of the PHP implies existence of certain objects with certain properties and shows us how to find them.

False - PHP only tells us the pigeons MUST exist in one of the holes. It does NOT specify which hole the pigeons are in. Think of the example of 367 students in the class implies 2 must have the same birthday. PHP does NOT specify which two students have the same birthday and/or which date that is.

Challenge: The number of ways to split 10 people into two 5-person teams to play volleyball is : 10!×10!/2 because forgetting the 2 in the denominator would result in an overcount by a factor 2, which can be interpreted as an additional assignment of a court to each team on which to play (not required by the problem!).

False, the correct answer is C(10,5)/2 (don't forget to divide by 2 because otherwise, we are counting ordered pairs of teams; or equivalently, choice of two teams and a side of the net to play, which is not asked by the problem).

A student can visit the office hours of any of the GSIs for the class as long as he/she has taken a quiz in that GSIs section.

False- A student may visit the office hours of any of the GSIs for the class, even without having taken a quiz in that GSI's section.

AxBxC for some sets A, B, and C is another set made of all possible triplets (x,y,z) where x, y, and z are any elements of the three sets.

False- By definition, A × B × C = {(x, y, z)|x ∈ A, y ∈ B, z ∈ C}. A × B × C is the set of all possible triplets (x,y,z) where x is an element of A, y is an element of B, and z is an element of C.

We can turn any counting problem into a problem using the product rule or the sum rule.

False- Even though these techniques are essential, we will see other techniques that have to be used for some counting problem.

To find how many natural numbers greater than or equal to n, are divisible by d, we calculate n/d and round up in order to not miss any numbers.

False- Round down, not up.

Among the problems we considered so far in class, a multi-stage process can be encoded (and solved) by either dependent choice or independent choice at each stage, split into mutually exclusive cases, or split into "good" and "bad" cases.

False- The principle of inclusion-exclusion can also be used to solve problems we've encountered so far.

To show that a conclusion does not follow from the given conditions, we need to do more work than just show one counterexample.

False- one counterexample is enough

Reversing the order of stages in a process does not affect the difficulty or efficiency of solving the problem.

False- to count the number of teams with a captain within a set of 11 people, it is much easier to consider the choice of the captain as the first stage.

The k-permutations of an n-element set are a special case of the k-combinations of this set.

False. Combinations and permutations are not special cases of each other. They are different objects.

The symmetry of permutations can be seen in the identity P(n,k)=P(n,n-k) for all integer n,k

False. Even though the analogous identity is true for combinations, it is false for permutations; e.g., P(4,3) = 4×3×2 ̸= P(4,4−3) = P(4,1) = 4.

Interpreting the same quantity in two different ways is not useful in proving binomial identities because, ultimately, one of the interpretations is harder (or impossible!) to calculate on its own.

False. Proving binomial identities through two interpretations is often the slickest way to do it.

(Deep!) The alternating sum of the numbers in an even-numbered row of Pascal's triangle is zero for the simple reason that Pascal's triangle is symmetric across a vertical line; but the same statement for an odd-numbered row requires some deeper analysis since the numbers there do not readily cancel each other.

False. The alternating sum of an odd-numbered row is zero because of symmetry.

The number of combinations C(n,k) is the number of permutations P(n,k) divided by the number of permutations P(n,n).

False. To obtain C(n,k), we divide P(n,k) by P(k,k). In fact, that is because in the case of combinations the order does not matter. And P(k,k) repre- sents the number of ordered arrangement we can realize with k elements.

A counterexample is a situation where the hypothesis (conditions) of a statement are satisfied but the conclusion is false.

True

Among the problems we considered in class, a multi-stage process can be encoded (and solved) by either dependent choice or independent choice at each stage, or split into mutually exclusive cases.

True

An identity is an equality that is always true for any allowable values of the variables appearing in the equality.

True

An ordered k-tuple can be thought of some permutation of k elements, while an unordered k-tuple can be thought of a combination of k elements (perhaps, coming from a larger set).

True

Challenge: To prove that there are some two points exactly 1 inch apart colored the same way on a canvas painted in black and white, it suffices to pick an equilateral triangle of side 1 in on this canvas and apply PHP to its vertices being the pigeons and the two colors (black and white) being the holes.

True

Counting problems where the phrase "at least once" appears may indicate using the complement, or equivalently, counting all cases and subtracting from them all "bad" cases.

True

In this class, it is hard to say which is more important: to know all versions of the Pigeonhole Principle or to have familiarized oneself with the syllabus in detail; but one thing is sure: both are indispensible to do well in the course.

True

One good reason for 0! to be defined as 1 is for the general formula with factorials for C(n,k) to also work for k=0.

True

The binomial coefficients first increase from left to right along a row in Pascal's triangle, but then they decrease from the middle to the end of the row.

True

The coefficient of x^3y^2 in (x + y)^6 is 0 because 2 + 3 does not equal 6; yet, it appears twice in the expanded form of (x + y)^5.

True

The product rule for counting usually applies if we use the word "AND" between the stages of the process, while the sum rule for counting is usually used when we can finish the whole process in different ways/algorithms and we use the word "OR" to move from one way to another.

True

Tree diagrams present a visual explanation of a situation, but unless one draw the full tree diagram to take into account all possible cases, the problem is not solved and will need more explanation/justification.

True

We can use the Binomial Theorem to prove all sorts of binomial identities, provided we recognize what x, y, and n to plug into it.

True

We did not prove de Morgan's Laws in class, however, one can either prove them as bonus HW or find their proof in the textbook or on the internet.

True

We solved in class the problem of finding the size of the power of a set by setting up a multi-stage process with 2 independent choices at each stage.

True

Challenge: Erdos-Szekeres Theorem on monotone sequences is a generalization of the class problem on existence of an increasing or a decreasing subsequence of a certain length, and its proof assumes that one of two possibilities is not happening and shows that the other possibility must then occur.

True - See proofs in class/book/handouts.

Proof by contradiction can be used to justify any version of the PHP.

True - See proofs in class/book/handouts.

We use 1 more than the ceiling (and not the floor) function in the statement of the Most General PHP because, roughly, we want to have one more pigeon that the ratio of pigeons to holes in order to "populate" a hole with the desired number of pigeons.

True - The ceiling function takes into account the uprounding to populate the extra pigeon(s).

A phrase of the type "at least these many objects" indicates what the pigeons should be in a solution with PHP, while "share this type of property" points to what the holes should be and how to decide to put a pigeon into a hole.

True - This is the general indicator. Still, phrasing the question into this type is the tricky part.

The basic combinatorial relation satisfied by binomial coefficients that makes it possible to identify all numbers in Pascal's triangle as some binomial coefficients can be written as n−1 + n−1 = n for n, k ≥ 1.

True.

The binomial coefficients appear in Pascal's triangle, as coefficients in algebraic formulas, and as combinations.

True.

Challenge: The formula 1+2+3+···+n = n+1C2 for n ≥ 1 is a special case of the Hockeystick Identity kCk + k+1Ck + k+2Ck+ · · · + nCk= n+1Ck+1 for n ≥ k ≥ 0.

True. Take k = 1 and then we get 1C1 + 2C1 + 3C1 +...nC1 = 1 + 2 + ...+ n = n+1C2 .

It is possible to use Calculus to prove combinatorial identities.

True. We can take the derivative of (1 + x)^n

Challenge: To prove some identity combinatorially roughly means to count the same quantity in two different ways and to equate the resulting expres- sions (or numbers).

True. We found the quantity of the two members of the identity and we show that they are equals.