Math 10B Midterm 1 True/False

The coefficient of x^3y^2 in (x+y)^6 is 0 because 2+3 ≠ 6; yet, it appears twice in the expanded form of (x+y)^5.

True

The formula 1 + 2 + 3 + ... + n = [(n+1) choose 2] for n ≥ 1 is a special case of the Hockeystick Identity for n ≥ k ≥ 0.

True

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

To prove the Probability "Baby P.I.E" property P(A ∪ B) = P(A) + P(B) − P(A ∩ B), one could split A ∪ B on the LHS into three disjoint subsets, similarly split on the RHS A and B each into two subsets, cancel, and match the resulting probabilities on the two sides.

True

To show that something is possible, it suffices to provide just one way of doing it, but to show that something is always true, we need to provide a proof that works for all cases.

True

We can prove the recursive formula for the Stirling numbers in a way very similar to the basic binomial identity by selecting one special object and discussing the two possible cases from its viewpoint.

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 need to add several Stirling numbers of the second kind in order to count the ways to distribute distinguishable balls to indistinguishable boxes because all situations split into cases according to how many boxes are actually non-empty.

True

When calculating the probability P(A) for some event A ⊆ Ω on an "equally likely" finite probability space (Ω, P), we can simply count the number of outcomes of A (the good possibilities) and divide that by all outcomes in Ω (all possibilities

True

When considering conditional probability, we are restricting the original outcome space Ω to a smaller subspace B given by the condition of something having happened.

True

Within the inductive step of a proof by MMI, we may occasionally need to use Sn−1, Sn−2, Sn−3, or some previous Sk (instead of Sn ) in order to prove Sn+1.

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: "AND" requires the succession of several stages that all need to occur, whereas "OR" offers exclusive options for the process.

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: Because with the formula, we would have C(n, 0) = n!/0!×(n−0)! that gives 1 if 0! = 1

The number of k−combinations from n elements with possible repetitions is [(n+k-1) choose (n-1)] and it matches the answer to the problem of distributing k identical biscuits to n hungry (distinguishable) dogs.

True: By the definition of k−combinations from n elements with possible repetition, n is the number of distinguishable boxes, and k is the number of indistinguishable objects.

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: Considering the elements of the set one by one, each stage of the process consists of including the element in the set or not. So there is two independent choices at each stage: either the element is included in the set, or it is not.

The property P(A ∪ B) = P(A) + P(B) − P(A ∩ B) for any A, B ⊆ Ω is true for any probability space (Ω, P), but it can be proven using baby P.I.E. only when P is the "equally likely" probability on a finite outcome space Ω, while a more general argument is needed for other (Ω, P).

True: Even though in the "equally likely" case we can use baby P.I.E. to prove this (because P.I.E. uses the sizes of the sets and so will we when we calculate P(A) for any event in an "equally likely" space), for probability space that is not equally likely, we cannot directly use |A|/|Ω| to calculate the probability of event A.

The equation x1 + x2 + x3 + x4 = 10 in natural numbers has as many solutions as trying to feed 4 (different) dogs with 6 (identical) biscuits.

True: If x1, x2, x3, x4 are natural numbers, then xi's will be at least 1. So this is a indistinguishable objects/distinguishable boxes/surjective problem.

When selecting at random two cards from a given 6-card hand (from a standard deck) that is known to contain 2 Kings, it is more likely to end up with at least one King than no King.

True: P(no king) = [4 choose 2]/[6 choose 2] = 6/15. P(at least 1 king) = 1 - 6/15 = 9/15.

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)

The number of 7-letter English words (meaningful or not, with possible repetition of letters) is not equal to the ways to distribute 7 equal bonuses to 26 people (with possible multiple-bonus winners).

True: The first question's answer is 26^7, whereas the second is [(7+26-1) choose 7].

Of the 12 possible balls-into-boxes problems, so far we found direct formulas for 10 problem types, even though 3 of these 10 formulas used summations; however, we didn't list direct formulas for 2 of the problem types because no one has succeeded in finding such formulas, no matter how deep and long they have tried these problems.

True: The formulas were not found in the case of identical balls and identical boxes, except for the injective case.

The equation x1+x2+x3+x4 = 10 in natural numbers where the order of the variables does not matter has as many solutions as the number of ways to split 10-tuplets (10 identical kids) into 4 identical playpens, where each playpen has at least one kid.

True: The order does not matter matters. x1, x2, x3, x4 are identical boxes.

If we do not know the final precise answer to a problem, we cannot apply MMI until we conjecture what this answer is.

True: The principle of MMI is to show that a statement is true. A statement to be clearly defined has to be an assertion and by consequence a (often mathematical) sentence that we can assign the value false or true to it.

In general, it is harder to handle balls-into-boxes problems where the function must be surjective than where the function is injective or there are no restrictions on it.

True: The surjective problem is harder than the function being injective or no restrictions (we omit the surjective problem in distinct/distinct part, and indistinct/distinct/surjective problem is based on the formula for indistinct/distinct/no restriction).

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: Unless you draw the full tree diagram (which is often too large to be drawn in a reasonable amount of time/space), you must adequately explain and justify your reasoning in words.

It may not be possible to calculate P(A ∩ B) using just P(A) and P(B), but if we also know one of P(A | B) or P(B | A), we can do it

True: Using the definition of conditional probabilities: P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B).

'Harry Potter is immortal.' is not suitable for a proof by MMI, but it can be paraphrased into such a suitable statement.

True: We can paraphrase it like the following statement: P(n): 'Harry Potter is alive on day n'. Then if we show the base case (that Harry is alive on day 1) and the inductive step : (that is if Harry is alive on some day n, he will be alive the next day), then by induction, we showed that Harry Potter is immortal.

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

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

When making the inductive hypothesis 'Suppose Sn is true.' we need to say 'for some n '; yet, we cannot specify a particular number for n here.

True: We have to consider a value of n within the range we want to prove our statement P(n) is true. Then, declaring that statement is true for some value of n, we are able to use it in our induction step.

In class, we defined an inversion as two adjacent elements in a list that are out of order.

False: An inversion is any pair of elements which are out of order.

MMI is not really necessary to formally prove the property: P(A1 ∪ A2 ∪ · · · ∪ An) = P(A1) + P(A2) + · · · + P(An) for pairwise disjoint Ai's because the property is quite intuitive and, to prove it, we can just apply over and over again the basic property of probabilities P(A ∪ B) = P(A) + P(B) for various non-overlapping A, B ⊆ Ω.

False: By applying the basic property P(A ∪ B) = P(A) + P(B) we are actually separating A1 ∪ A2 ∪ · · · ∪ An to be (A1 ∪ A2 ∪ · · · ∪ An−1) ∪ An, in which we would also need to prove the formula for n − 1 case. So it actually follows the idea of MMI. The phrase "apply over and over again" something should tell you that, most likely, induction is close by and should be used to formalize the solution.

A × B × C 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.

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.

It is incorrect to say that the elements of Ω are "outcomes" since they are actually inputs of the probability P, and not outputs.

False: Even though the elements of Ω are inputs for the probability function P, they are also outcomes of experiments, and hence we have the name "outcome space" for Ω. So we can call the elements of Ω "outcomes" since they are indeed the outcomes of some random measurements or experiments.

It is always true that ⌊ x ⌋ ≤ x ≤ ⌈ x ⌉ for any real number x, but equality of the two extreme terms of this inequality is never possible.

False: Even though the inequalities is TRUE for all real numbers, the latter part of the statement is FALSE. Equalities hold for all integers. For a counter-example: 0 is a real number, but ⌊ 0 ⌋ = 0 = ⌈ 0 ⌉

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.

An algorithm is a finite sequence of well-defined steps that always lead to the correct (desired) output.

False: Formally, an algorithm does not always lead to the desired output. However, a *good* algorithm (the ones we are interested in and as discussed in class) is a finite sequence of well-defined steps that always leads to the correct (desired) outcome for all input data of a certain type in a finite amount of time.

P(A | B) can never be equal to P(B | A) unless P(A) = P(B).

False: If P(A ∩ B) = 0, then P(A | B) = P(B | A) = 0, but you can P(A) ≠ P(B). For instance, when rolling a die, consider the events "obtaining 1" and "obtaining 2 or 3".

The stable marriage algorithm produces the same final stable pairing, even if we reorder the "men" or if we switch the places of "men" and "women".

False: In both situations, the algorithm might produce a different yet still stable set of pairings.

It is never possible to pair up 2018 people into stable roommate pairs because, even if we manage to pair up 2014 of them in stable pairs, there will always be 4 of them to produce a counterexample of an impossible stable pairing.

False: It depends on everyone's preference list. The counterexample we gave in class for an impossible stable roommate matching for 4 people does not mean that it will always occur in any situation. Not only stable matchings are possible, but also *happy* matchings are possible: it all depends on the particular preference lists of all people involved.

It is never necessary to show the first several base cases in a proof by MMI; indeed, we do this just to boost our confidence in the truthfulness of the statement of the problem and we need to show only that the first base case is true.

False: It depends on how many previous statements you need to prove the inductive step. For example you want to show that P(n) is true for any positive integer n by induction. During the inductive step, for some k greater than or equal to 2, you will want to prove that P(k + 1) is true. In some (challenge) cases, you might need not only to use the fact that P(k) is true to prove that P(k + 1) is true, but sometimes also that P(k − 1) is true. Then, the inductive step becomes: P(k − 1), P(k) ⇒ P(k + 1). That means that in the base case you will have to show that P(1) and P(2) are true.

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.

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.

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.

An argument by contradiction can be avoided if we are careful not to make mistakes in our proof.

False: Reaching a contradiction is actually *good* within a proof by contradiction, and it is indeed our goal to reach a contradiction when performing such a proof, not to avoid it.

To find how many natural numbers ≤ n are divisible by d we calculate the fraction (n/d) and round up in order not to miss any numbers.

False: Round down, not up.

Within MMI, the inductive step 'If Sn is true then Sn+1 is also true.' implies that Sn+1 is true.

False: Sn+1 is true only if Sn is true but no where we are told that Sn is true.

The Quick Sort algorithm is, on the average, faster than the Bubble Sort algorithm because the number of inversions in the list being sorted increases faster during the Quick Sort algorithm.

False: The Quick Sort algorithm is, on the average, faster than the Bubble algorithm but the reasons is that the number of inversions in the Quick Sort algorithm decreases (not increases) faster.

The Stirling number of the second kind S(n, j) counts in how many ways we can distribute n (distinguishable) kids into j identical playpens so that no kid is left outside of a playpen.

False: The Stirling number of the second kind S(n, j) counts in how many ways we can distribute n (distinguishable) kids into j identical playpens so that no playpen is left empty.

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. Note that an odd numbered row has an even number of elements.

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).

The probability function P is defined as P : Ω → [0, 1] such that P(Ω) = 1, P(∅) = 0, and P(A ∪ B) = P(A) + P(B) for any disjoint subsets A and B of Ω.

False: The domain of probability function P should be the power set of Ω, i.e. the set of all subsets of Ω. The rest of the question is true.

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

Even though the analogous identity is true for combinations, it is false for permutations; e.g., P(4, 3) = 4×3×2 6= P(4, 4−3) = P(4, 1) = 4.

The expression (x + y + z + t)^2018 has [(2020) choose 3] terms after multiplying through but before combining similar terms, and 4^2018 terms after combining similar terms.

False: The expanded expression of (x + y + z + t)^2018 has 4^2018 terms after multiplying through but before combining the similar terms; after combining the similar terms, the expression consists of terms in the following format: (x^p)(y^q)(z^r)(t^s), where p + q + r + s = 2018. 2018. So the number of terms is just the number of nonnegative integer solutions of p + q + r + s = 2018, which is [(2018+4-1) choose (4-1)].

The number of roommate pairings among 2018 people can be written as 2017 · 2015 · 2013 · · · 3 · 1 , or alternatively also as (2018!)/(2^2018·1009!)

False: The first expression is correct, but the second should be (2018!)/(2^1009·1009!)

The number of ways to distribute b distinguishable balls into u distinguishable urns is u!S(b, u) and the answer was obtained by solving first the same problem with indistinguishable urns and then labeling (or coloring) the urns to make them distinguishable.

False: The formulas S(b, u) and u!S(b, u) are for the case of surjective functions (not just any functions).

The maximal possible number of inversions in a list of 10 numbers is P(10, 2) = 90, while the minimal such number is 1.

False: The maximal possible number of inversions in a list of 10 numbers is C(10, 2) = 45, while the minimal such number is 0.

At the end of a successful application of MMI, we conclude that Sn is true for some particular n's.

False: The principle of MMI is to actually prove that a statement P(n) is true for any value n within some range.

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.

The formula P(A) = P(A | B)P(B) + P(A | Bcomplement)P(Bcomplement) works for any events A, B ⊆ Ω as long as P(B) > 0.

False: This formula requires P(B) < 1 as well.

The formula ⌈ N/d ⌉ appears when calculating the probability of a natural number n ≤ N to be divisible by d.

False: To calculate the probability of a natural number n ≤ N to be divisible by d, we need to determine how many natural numbers n ≤ N are divisible by d, which is ⌊ N/d ⌋. And the probability will be ⌊ N/d ⌋/N.

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) represents the number of ordered arrangement we can realize with k elements.

Two problem types in the 12-fold way table have extremely simple answers because an injective function cannot have a smaller domain than co-domain.

False: Two problem types in the 12-fold way table have extremely simple answers because an injective function cannot have a LARGER domain than co-domain.

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.

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

True

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

A stable matching between n jobs and n people means that even if some person A prefers a job B on this list of jobs that he has not been given, the company that offers job B has hired another person C on it that they prefer to person A.

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

Bonferroni's inequality P(E ∩ F) ≥ P(E) + P(F) − 1 is, in disguise, the well-known fact that P(E ∪ F) ≤ 1.

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

Despite the suggestive notation, the conditional probability P(A | B) was originally defined through a formula and we had to prove that it indeed is in [0, 1] in order to consider P(A | B) as an actual probability.

True

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

In general, it is harder to handle balls-into-boxes problems where the boxes are indistinguishable than where the boxes are distinguishable.

True

It is possible to use Calculus to prove combinatorial identities.

True

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

True

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) choose (k-1)] + [(n-1) choose k] = [n choose k] for n,k ≥ 1.

True

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

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

If we experiment with throwing two fair dice and adding up the two values on the dice, and if we decide to represent the outcome space Ω as the set all of possibilities for the sum; i.e., Ω = {2, 3, . . . , 12}, then the corresponding probability P will not be the "equally likely" probability, making us reconsider the choice of the outcome space Ω in the first place.

True: We know there is only one pair that can give sum 2 ((1, 1)), but there are 3 pairs that give sum 4 ((1, 3),(2, 2),(3, 1)), so the probability of occurrence of 2 and 4 is not equal. So this is not an "equally likely" probability space if we consider the outcome space as the set of all possible sums. Thus, if we can create an "equally likely" probability space, we may be able to simplify our calculations. Such a space consists of all possible *pairs* of outcomes; e.g., {(i, j)|for i = 1, 2, ..., 6, j = 1, 2, ..., 6}.

When A ⊆ B, the conditional probability P(A | B) can be expressed as the fraction P(A)/P(B) (given all involved quantities are well-defined).

True: When A ⊆ B, A ∩ B = A.