1. Sets, Combinatorics & Probability

The following constitutes a basic introduction to sets, combinatorics, and probability necessary for competent statistical analysis. The examples we use will be heavy on genetics because of our own research interests. Read this chapter carefully and try the problems; then use it as a reference for the rest of the course.

1.1. Sets

We introduce the idea of a set here only because it is often helpful to think of events as sets, and it is of events that we calculate probabilities. For example, from a mating of type Aa × Aa there are three possible offspring: AA, Aa and aa–a set, or sample space, of three elements. Any event concerning this one mating can be expressed as a subset of this set of three primitive events, e.g., the event "A_" is represented by the set (AA, Aa), the event "homozygote" is represented by (AA, aa). In practically all of the problems we encounter, the sample space will consist of a finite number of primitive events, and any event we can imagine will correspond to a subset of that sample space.

Examples 1.1

A. Suppose the mating Aa × Aa produces three children. Then the sample space consists of 27 events. Note, here, the primitive events are themselves sets.   

AA,AA,AAAA,Aa,AAAA,aa,AAAa,AA,AA
Aa,Aa,AAAa,aa,AAaa,AA,AAaa,Aa,AA
aa,aa,AAAA,AA,AaAA,Aa,AaAA,aa,Aa
Aa,AA,AaAa,Aa,AaAa,aa,Aaaa,AA,Aa
aa,Aa,Aaaa,aa,AaAA,AA,aaAA,Aa,aa
AA,aa,aaAa,AA,aaAa,Aa,aaAa,aa,aa
aa,AA,aaaa,Aa,aaaa,aa,aa

The event "at least two recessive children" is the set {aa,aa,AA; aa,AA,aa; AA,aa,aa; aa,aa,Aa; aa,Aa,aa; Aa,aa,aa; aa,aa,aa}. The event "the first child is AA" comprises nine of the primitive events, and the event "the second child is a homozygote" comprises 18 of the primitive events. Any event imaginable may be represented by some subset of these 27 primitive events. How many are there? Answer: 227-1=134,217,727.

B. Consider a sequence of six bases (nucleotides). There are 46 = 4096 such sequences (we won't display them!). The event "AANNTT", where N denotes any base, represents a set comprising 16 of these sequences.

1.1.1. Set.

A set is simply a collection of objects. No order is implied. An object included in a set is said to belong to that set and to be an element of the set. The elements of a set can be anything, even other sets.

Example 1.2

Let the set A = {Mike, the moon, a particular loaf of bread, the set of integers greater than 1}. This set contains four elements, one of which is, itself, a set containing a countable infinity of elements.

There are two ways of defining a particular set. The first is simply to enumerate its elements as we did for set A, above. The second is to characterize the elements in some way, for example, let B = {x| x is a female}. This statement is read: "B consists of all organisms, x, such that x is a female." That is, B is the set of all females.

1.1.2. Union.

The union of two sets, A and B, written A ∪ B, is the set of all elements belonging either to A or to B or to both.

Examples 1.3

A. The union of the set {a,b,c} and the set {b,c,d,e} is {a,b,c,d,e}.

B. Consider the mating Aa × aa. The set of all possible families of size three is X ={(Aa,Aa,Aa), (Aa,Aa,aa), (Aa,aa,Aa), (aa,Aa,Aa), (Aa,aa,aa), (aa,Aa,aa), (aa,aa,Aa), (aa,aa,aa)}. The set of all such families in which the first child is aa is Y = {(aa,Aa,Aa), (aa,Aa,aa), (aa,aa,Aa), (aa,aa,aa)} and the set of all such families in which the first two children are aa is Z={(aa,aa,Aa), (aa,aa,aa)}. In this case, the union of Y and Z is just Y.

C. Considering our six-base sequences, again—the set "AACNTT" = (AACCTT, AACGTT, AACATT, AACTTT) and the set "AANCTT" = (AACCTT, AAGCTT, AAACTT, AATCTT). The union "AACNTT" ∪ "AANCTT" = (AACCTT, AACGTT, AACATT, AACTTT, AAGCTT, AAACTT, AATCTT).

Note that union corresponds to the English word "or" in the inclusive sense. Throughout this book, we will usually write "A or B" rather than "A ∪ B" when A and B refer to events.

1.1.3. Complement.

The complement of a set, A, written Ac, is the set of all elements not belonging to A. Our use of complement arises from the helpful fact that it is sometimes the case that to find the probability of an event, A, it is easier to calculate [1 - (the probability of the complement of A)] than the probability of A, itself.

Examples 1.4

A. The P{of at least one girl in a family of five children} = 1 - P{of no girls in a family of five children}.

B. In families of size six from the mating Aa x aa, the event "at least two aa children" (the union of four events) is the complement of the event "less than two aa children" (the union of two events).

C. For our six-base sequences, the complement of the set "ANNNNN" is the set (CNNNNN, GNNNNN, TNNNNN).

1.1.4. Intersection.

The intersection of two sets, A and B, written A ∩ B, is the set of all elements belonging to both A and B.

Examples 1.5

A. The intersection of the sets N={b,c,d,e,g} and P={a,c,e,g,h} is the set N∩P={c,e,g}.

B. The intersection of the sets Y and Z, in Example 1.3.B, is Z.

C. The intersection of the sets M={x|x is a male} and F={x|x is a female} is empty.

D. The set "AACNTT" ∩ "AANCTT" = the set (AACCTT).

Note that intersection corresponds to the English word "and". We will often write "A and B" or just "AB" to denote intersection.

1.1.5. Subset.

A set, A, is said to be a subset of the set, B, written A ⊆ B, if every element of A is also an element of B. Z was a subset of Y, in the Example 1.3.B.

1.1.6. Some properties of union and intersection.

For any three sets, A, B and C:

  AB = BA A(BC) = (AB)C  
  A ∪ B = B ∪ A A ∪ (B ∪ C) = (A ∪ B) ∪ C
  A ∩ (B ∪ C) = AB ∪ AC (A ∪ B) ∩ (A ∪ C) = A ∪ BC
DeMorgan's Rules:
  (A ∪ B)c = (Ac)(Bc) (AB)c = (Ac) ∪ (Bc)

The operations of union and intersection have some of the properties of addition and multiplication. Note certain peculiarities, however. For example, A ∩ (A ∪ B) = A ∪ AB = A. You can convince yourself of the truth of these various equalities by studying the Venn diagram, below.

The rules also apply to the truth value of statements. For example, if you let A = "physicians are wise" and B = "professors are stupid", and so forth, all the rules apply.

1.2. Combinatorics

To calculate probabilities we will often be required to determine all possible results of a particular experiment. The following "rules of counting" will be helpful.

1.2.1. Rule of sum.

If event X can happen in x different ways and a distinct event Y can happen in y different ways, then the event "X or Y" can happen in (x + y) ways.

Examples 1.6

A. For an X-linked locus with two alleles, a woman can be any of three genotypes, a man any of two. A random individual (sex unspecified) can be any of five genotypes.

B. The event "AANTTT" or "AAANTT" can happen in 7 ways. Why doesn't the rule work here?

C. A menu lists six entrees, five desserts, and three salads. How many ways can I order an entree or a salad? Answer:  6+3.

1.2.2. Rule of product.

If event X can happen in x ways and a distinct event Y can happen in y ways, then the event "X and Y" can happen in xy ways.

Example 1.7

For an X-linked locus with two alleles, a woman can be any of three genotypes, a man any of two. A couple can then be any of six pairs of genotypes. This is the "menu" rule: e.g., if there are 6 entrees and 5 desserts, there are 30 possible meals. Notice that this extends to more than two selections. If there are, in addition, 7 appetizers to choose from, the total number of different meals is 210.

1.2.3. Power set.

Any set of n elements has 2n subsets. The set of all subsets of a set is called its power set. A power set thus has 2n elements.

Example 1.8

If A = {a,b,c}, the power set of A is the set { ∅, a, b, c, ab, ac, bc, abc}. "∅", here, denotes the empty set. It is a convention that the power set includes as one of its elements the empty set, the set containing no element. An example: How many sounds can you make on a piano with 88 keys? Answer: 288-1. We don't count the empty set here because it represents no sound!

1.2.4. Permutations.

A permutation is an ordered arrangement. The number of permutations of n distinct objects is

n! = n(n-1)(n-2)(n-3)…(2)(1).

Examples 1.9

A. The number of permutations of the three letters, a, b and c, is six: abc, acb, bac, bca, cab, cba.

B. Consider a chromosome containing five loci, A, B, C, D, and E, on its long arm. There are 5! = 120 possible maps (orders) for these loci.

1.2.5. Ordered selections.

The number of ordered selections of r things out of n distinct objects is

Examples 1.10

A. The number of ordered pairs that can be gotten from the set {a, b, c} is six: ab, ba, ac, ca, bc, cb.

B. Eight students wish to see you next Thursday, but you have only five times free. The number of different appointment schedules possible for that day, assuming you use all your time slots, is 8×7×6×5×4 = 6720.

1.2.6. Unordered selections (binomial coefficients).

The number of ways r things can be chosen out of n distinct objects (regardless of order) is

These are sometimes called "combination numbers" because they give the number of combinations (unordered selections) of size r that can be made from a set of size n.

Examples 1.11

A. If there are n alleles at a locus, the number of heterozygotes possible is .

The number of homozygotes is n, so the total number of genotypes is n(n-1)/2 + n = n(n+1)/2.

B. With respect to the scheduling problem in Example 1.10, you can pick five of the eight students to see on Thursday in ways and given the chosen five, there are 5! = 120 ways to schedule them. The product is 56×120 = 6720, as before.

C. In how many ways can an RNA molecule of 10 bases be constructed with 3 U, 3 G, 3 A, and 1 C?

Answer:  

Note that , , and , which lead to the familiar Pascal's triangle






1








1
1






1
2
1




1
3
3
1


1
4
6
4
1
1
5
10
10
5
1










1.2.7. Multinomial coefficients.

The number of ways that a set of n objects can be divided into k distinguishable parts of which the first contains r1 objects, the second r2 objects, and so on (r1+r2+…+rk = n) is

Example 1.12

From a mating of type Aa×Aa, four children are produced: 2 AA, 1 Aa and 1 aa. How many different birth orders are possible? The answer is 4!/2!1!1! = 12. Let 1=AA, 2=Aa and 3=aa. The 12 orders are: 1123, 1132, 1213, 1312, 1231, 1321, 2113, 3112, 2131, 3121, 2311, 3211.

1.3. Probability

One can often think of probabilities as proportions. For example, to ask "what is the probability that a randomly selected person is a heterozygote?" is simply to ask "what proportion of all persons are heterozygotes?" Often a probability is thought of as the proportion of times an event will occur in the long run, in a series of repeated "experiments". For example, consider the simple pedigree below.

A normal man whose father is an albino marries an albino woman. What is the risk of albinism to their child? The prospective father is definitely a heterozygote. The risk to the child is therefore one-half. By this we mean that in a large series of families just like this one, each producing a child, in the long run about half of the children will be albino and half will be normal. We are forced to think of this particular family as a member of a larger group of families in which the proportion of affected children is expected to be one-half. The larger group may not even exist except in our imagination. If it did exist, and we told each of the families that their child will be albino, we would be correct half the time and incorrect the other half in the long run.

The above view of probability is not without its problems. The concept of in the long run is not easy to define in a precise manner. One version of this approach is to say that the probability of an event, E, is said to be p, if as the sample size, n, gets larger the observed proportion of cases of E, pE, approaches p in the sense that

Pr(|pE-p| > e) → 0 for any e as n → ∞.

This is, of course, hopelessly circular, because we are assuming the concept of probability to define it.

The interpretation of probability described above is usually called objective probability or frequentist probability. An alternative interpretation is that probability is simply a measure of belief; this is known as subjective or personal probability. For example, suppose I claim that the probability that Hillary Clinton will run for President in 2016 is 0.6. That is an expression of my conviction that she will run. It is difficult—perhaps impossible—to think of Clinton embedded in a series of possible election campaigns among which she runs 60% of times and demurs 40%.

1.3.1. Probabilities are normed.

For any event, E, 0 ≤ P(E) ≤ 1.

Example 1.13

An event with Pr = 1 must happen. An event with Pr = 0 cannot happen.

1.3.2. The Pr of the union of two events.

For any two events, E1 and E2, the probability of their union is

P(E1 or E2) = P(E1) + P(E2)-P(E1E2).

Example 1.14

Consider a mating of type Aa × aa. Let E1 be the event that the first child is aa and let E2 be the event that the second child is aa. What is the probability of the event "E1 or E2", that either child is aa? It is

P(E1) + P(E2) - P(E1E2) = 1/2 + 1/2 - 1/4 = 3/4.

Note that this is 1-P(neither is aa) = 1-P(both are Aa)

                                                = P(at least one is aa).

There are often multiple ways of expressing the same event.

If two events, E1 and E2, are mutually exclusive, that is, their intersection is empty (E1E2 = ∅) and therefore the Pr of their intersection is zero (P[E1E2] = 0), then we have

P(E1 or E2) = P(E1) + P(E2), as a special case.

Example 1.15

The probability that a child from the mating Aa × Aa is either AA or Aa is, from our general rule: P(AA or Aa)= P(AA) + P(Aa) - P(AA and Aa); the last term is clearly zero because the two events AA and Aa are mutually exclusive (also called 'disjoint'). A child cannot be both. So P(AA or Aa) = 1/4 + 1/2 = 3/4.

1.3.3. The Pr of the union of more than two events.

Let Σ denote summation. The union of n events is

P(E1 or E2 or … or En) = ΣP(Ei) - ΣP(EiEj) + ΣP(EiEjEk) - … -(-1)nP(E1E2…En).

For three events we have

P(A or B or C) = P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC)

This formula can easily be understood by reference to the Venn diagram in section 1.1.6.

Example 1.16

Consider, again, the Aa × aa mating. What is the probability that either the first child or the second child or the third child is aa? Let A= first child is aa, B=second child is aa, and C=third child is aa. Then P(A or B or C) = 1/2 + 1/2 + 1/2 - 1/4 - 1/4 - 1/4 + 1/8 = 7/8. This is, of course, 1 - P(all are Aa).

1.3.4. Conditional probability.

The conditional probability of E1 given E2 is defined as

In words, we are asking: In those cases in which E2 has occurred, in what proportion has E1 also occurred?

Examples 1.17

A. From the mating Aa × Aa a child is born with the dominant phenotype. What is the Pr the child is Aa?

P(Aa|A_) = P(Aa and A_)/P(A_)

  = P(Aa)/P(A_) = (1/2)/(3/4) = 2/3.

B. If the population is in Hardy-Weinberg equilibrium (i.e., randomly mating), what proportion of A_ individuals are heterozygotes?

Answer: where q is the frequency of the a allele.

C. What is the Pr that a random oligo containing five bases has 2 U's given that it has at least one?

Answer:  

1.3.5. Writing an intersection in terms of conditionals.

This rule is sometimes very helpful in calculating probabilities. The following is true for any set of events E1, E2, … , En.

P(E1E2…En) = P(E1)P(E2|E1)P(E3|E1E2) … P(En|E1E2…En-1).

Here, the numbering of the events is arbitrary; any permutation of the subscripts (1,2,…,n) will work as well.

Examples 1.18

A. Suppose A = male, B = unfaithful, C = married. We can write

  P(ABC) = P(A)P(B|A)P(C|AB) = P(A)P(C|A)P(B|AC)

    = P(B)P(A|B)P(C|AB) = P(B)P(C|B)P(A|BC)

    = P(C)P(A|C)P(B|AC) = P(C)P(B|C)P(A|BC).

Think about the meaning of each expression in words.

B. Another example more germane to human genetics: Suppose you want to know the P(parents are both Aa | child is aa). From the definition of conditional probability we may write this as P(parents are both Aa and the child is aa)/P(child is aa). The numerator can then be written as P(parents are both Aa)P(child is aa | parents are both Aa), by our rule, which is easier to calculate because it involves a simple Mendelian segregation frequency (1/4). Thus, if the frequency of the A gene is p=1-q and we assume Hardy-Weinberg equilibrium we get

1.3.6. Independence.

Two events E1 and E2 are said to be independent if P(E1E2) = P(E1)P(E2) or what amounts to the same thing, if P(E1|E2) = P(E1). The two most common mistakes made when calculating probabilities are 1) calculating a conditional Pr with the wrong condition and 2) assuming two events are independent when they are not. Whenever you multiply two probabilities, ask yourself if the assumption of independence is justified.

Examples 1.19

A. The probability that the first and second children of the mating Aa×Aa are AA and Aa, respectively, is given by P(AA,Aa) = P(AA)P(Aa) = (1/4)(1/2). The assumption of independence made here is clearly justified.

B. Consider the following more subtle example: The mating is now A_ × Aa, that is, we do not know the genotype of the first parent, only the phenotype. Now, regarding the two children, P(AA,Aa) is not necessarily equal to P(AA)P(Aa). However, we can always write P(AA,Aa) = P(AA)P(Aa|AA). The reason the two children are no longer independent is because knowing the genotype of one of them tells us something about the genotype of the first parent which, thus, affects what we know about the other child.

C. What is the probability that a random RNA molecule of length 10 bases has at least one U? Assume A, C, G, and U are equally likely and different sites are independent. Answer: 1-(3/4)10

1.3.7. Partitioning an event.

Suppose E1, E2, …, En are mutually exclusive and exhaustive (i.e., all intersections are empty and the sum of the probabilities of the n events equals one). Then for any event, E, we can write

P(E) = ΣP(EEi) = ΣP(Ei)P(E|Ei)

where the sum is over i=1, 2, …, n.

Example 1.20

Let A = {x|x is male} and B = {x|x is over 50 years old}. Clearly A = AB or ABc and the events "AB" and "ABc" are mutually exclusive. (Again, draw a Venn diagram, if this is not clear to you.) Hence P(A) = P(AB) + P(ABc). We have simply partitioned males into those over 50 and those ≤ 50.

1.3.8. Bayes' theorem.

From 1.3.5 and 1.3.7, we see immediately that

This relation is called Bayes' Theorem after the Reverend Thomas Bayes, who first proposed it. Here P(Ei) is referred to as the prior probability of Ei. P(Ei|E) is the posterior probability of Ei. P(E|Ei) is called the likelihood.

Examples 1.21

A. Suppose you are at a party. One of your friends comes up to you and tells you that she has discovered a 95% accurate test for a certain kind of rare liver cancer (population frequency of 1 per million). The test, if it works, would be an important aid to early diagnosis. You are impressed. However, you start to think about her claim and ask, "What do you mean when you say your test is 95% accurate?" She responds immediately, "I mean that if you have the cancer, the test will be positive 95% of the time, and if you do not have the cancer, the test will be negative 95% of the time. She goes on describing the test and finally asks, "Would you like to come to my lab and take the test?" You see no harm and agree. The next day you have your blood drawn and the test is performed. That evening your friend calls to tell you the bad news; your test was positive. Your mind suddenly clears, and you realize there is a question, the answer to which now seems very important to you: Given that I have come up positive on this test, what is the probability that I have liver cancer?

What do you know? You have four items of data.

  P(C) = 0.000001, the frequency of this cancer
                     is one in a million,

  P(+|C) = 0.95, if you have the cancer,
                     the test will say so 95% of the time,

  P(-|notC) = 0.95, if you are healthy, the test will
                     say so 95% of the time,

  You came up positive.

What do you want to know?

Clearly, P(C|+) = ?, What is the Pr that you have the cancer given you are positive? This is exactly the kind of question that Bayes' theorem answers.

We have

  

B. Suppose a woman with an albino brother and normal parents marries an albino man. Her first child is normal. What is the Pr that her second child will also be normal? Let A = "mother is heterozygote" and B = "first child is normal". Then we have

  

Thus the probability that the mother is Aa given that she has had one normal child is

  

Hence the risk to her next child is (1/2)(1/2) = 1/4.

The application of Bayes' theorem is important for understanding diagnostic or biomarker testing, as described in Example 1.21.A. However, the terminology used in those settings differs from that in our discussion above. Using the notation in the example, P(C) is the prevalence of the disease (the prior probability). The sensitivity of the diagnostic test is P(+|C), and its specificity is P(-|notC). We can use Bayes' theorem and the prevalence, sensitivity, and specificity to estimate the positive predictive value (PPV) as P(C|+) and the negative predictive value P(notC|-).

In Example 1.21.B, you will notice that the crux of Bayes' theorem is simply that P(Ei|E) is proportional to P(Ei)P(E|Ei). Consequently, we can conveniently work in odds by ignoring denominators until the end. We will often use the following format in setting up problems utilizing Bayes' theorem:

Event E1 E2 En
prior Pr P(E1) P(E2) P(En)
likelihood P(E|E1) P(E|E2) P(E|En)
posterior odds P(E1)P(E|E1) P(E2)P(E|E2) P(En)P(E|En)

We may now get the posterior probabilities of E1 … En simply by dividing each of the quantities in the "posterior odds" row by their total. In our previous example the calculations would look like this:

Event Mother Aa Mother AA
prior Pr 2/3 1/3
likelihood 1/2 1
posterior odds 1/3 1/3

By prior Pr, here, we mean the probability that the mother is a heterozygote before we know that her child is normal, that is, prior to any knowledge about the child. The likelihood is the probability of having a normal child if the mother is Aa (or AA). The posterior odds are then proportional to the product of the prior probability and the likelihood. The posterior probability that the mother is Aa is then 1/3 ÷ 2/3 = 1/2.

1.4. Sample Problems

  1. The first philosopher states: "Either women are wise or all men are fools." The second philosopher replies: "I deny that women are fools and men are wise." Are these two philosophers agreeing or disagreeing with each other? Hint: Use DeMorgan's rules.
  2. Two committees were asked to look into the possible mishandling of a genetic counseling case by three counselors, A, B and C. The first committee concluded: "Either A and B were correct or A was correct and C was wrong." The second committee concluded "Either A was wrong or, on the other hand, B was wrong and C was right." Can both committees be correct in their findings?
  3. In a recent court case, a man accused a woman of having his child and asked the court to grant him custody of the infant. The woman denied that the child is his. The man, woman and child were tested using several genetic systems. What are all the conclusions that the court could reach regarding this case? Hint: There are five.
  4. Five pregnant women, each over 40 years of age, have been advised by their physicians to have amniocentesis performed. Each woman must decide first whether to have the procedure, and then if the procedure is performed and the fetus is abnormal, whether to abort or not. How many sets of decisions are possible?
  5. In the pedigree below for X-linked hemophilia in which a woman, X, has two affected brothers and an unaffected son, what is the probability that X is a carrier? Use Bayes' theorem.

  1. A sibship of size 3 consists of 1 MM, 1 MN, and 1 NN child. How many birth orders are possible?
  2. A sibship of size 6 consists of 2 AA, 3 Aa, and 1 aa. How many birth orders are possible? Is this question completely unambiguous?
  3. From the mating Aa × Aa, what is the Pr that the first child will either be A_ or a heterozygote? Does rule 1.3.2 apply here?
  4. Consider three autosomal loci, each with two alleles: (A,a), (B,b), (C,c). In a certain population, the frequencies of the eight possible gametes are:  ABC 0.05, ABc 0.10, AbC 0.05, Abc 0.30, aBC 0.10, aBc 0.05, abC 0.10 and abc 0.25. We pick a gamete at random. If it contains an A allele, we say the event "A" has occurred, if it contains b and C, we say the event "bC" has occurred, and so on.
    a. Are the events "A" and "B" independent in this population?
    b. Are "B" and "c" independent?
    c. What is P(A|B)?
    d. What is P(Ac|B)?
    e. What is P(C|ab)?
    f. Are the events "a|B" and "C|B" independent?
  5. From the mating MN AB × MM OO, what is the Pr that the first child is either MN or AO or MMBO?
  6. Here is a challenging but useful set of problems to test your understanding of the Hardy-Weinberg rule. The diagram below depicts the sample space generated by choosing two parents and a child at random from a population in Hardy-Weinberg equilibrium for a locus with two alleles, M and N, with frequencies, p and q, respectively. There are 27 "conceivable" genotype triples (some have probability zero).

    The 3 × 3 × 3 conceivable families with one child when each member is classified as MM, MN or NN. Assume random mating.



    a. Fill in the cells of the diagram including marginals (some have been filled in to get you started). Letting m=mother, f=father, c=child.
    b. What is the Pr that f is MM, m is MM and c is MM?
    c. What is the Pr that f is MN, m is MN and c is NN?
    d. What is the Pr that c is NN given that f is MN and m is MN?
    e. What is the Pr that f is MN given that m is MN and c is NN?
    f. Calculate Pr[f is MN and m is MN | c is NN).
    g. Calculate Pr[f = MN, m=MN | c = MM or MN).
    h. Calculate Pr[f=MM | m=MN, c=MN).
    i. Calculate Pr[f=MM | m=MM, c=NN).
  7. How many RNA chains of length n are there with exactly two U's?
  8. A polypeptide of 20 amino acids has 5 histidines, 6 arginines, 4 glycines, 1 asparagine, 3 lysines, and 1 glutamic acid. How many such chains are there? What is your answer if you are told the polypeptide begins with a histidine and ends with an arginine?