Combinatorics: Learning Objectives 1, 2, 3, 4

Part (a) describes a combination. Part (b) describes a permutation. For part (c), consider how many options there are for each of the r choices.

1. $ \binom{n}{r} = \frac{n!}{r!(n-r)!} $
2. $ P(n,r) = \frac{n!}{(n-r)!} $
3. $ n^r $

1. This scenario describes a combination, as the order of selection does not matter. The formula is "n choose r".
2. This scenario describes a permutation, as the order of selection is important.
3. In this scenario, for each of the $r$ choices, there are $n$ possibilities because the items are replaced. This is an application of the multiplication principle.

The key question is: "Does the order of selection matter?" If selecting person A then B is a different outcome from selecting B then A, it's a permutation. If it's the same outcome, it's a combination.

1. Permutation
2. Combination
3. Permutation (with replacement)
4. Combination

1. Permutation. The roles of president and vice president are distinct. Choosing Member A as president and Member B as vice president is a different outcome from choosing B as president and A as vice president. Order matters.
2. Combination. The group of three fruits is the same regardless of the order in which they were picked. Choosing an apple, banana, and cherry is the same outcome as choosing a cherry, banana, and apple. Order does not matter.
3. Permutation (with replacement). The sequence of answers creates a unique submission. Answering A for question 1 and B for question 2 is different from answering B for question 1 and A for question 2. Order matters.
4. Combination. The three internship positions are identical. One group of three selected interns (Alice, Bob, Carol) is identical to (Carol, Alice, Bob). Order does not matter.

Start with the factorial of the total number of letters. If there are repeated letters, you must divide by the factorial of the count of each repeated letter to eliminate overcounting.

1. $5! = 120$
2. $\frac{5!}{2!} = 60$
3. $\frac{10!}{3!3!2!} = 50,400$

1. "great": There are 5 distinct letters. The number of permutations is $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
2. "greet": There are 5 letters in total, but the letter 'e' is repeated twice. To find the number of distinct permutations, we calculate $\frac{5!}{2!} = \frac{120}{2} = 60$.
3. "statistics": There are 10 letters in total. The letter 's' is repeated 3 times, 't' is repeated 3 times, and 'i' is repeated 2 times. The number of distinct permutations is $\frac{10!}{3!3!2!} = \frac{3,628,800}{(6)(6)(2)} = \frac{3,628,800}{72} = 50,400$.

Fix an 's' at the beginning and an 's' at the end. This leaves 8 letters in the middle to be arranged. Count the repetitions among these 8 middle letters and use the formula for permutations with repeated items.

$\frac{8!}{3!2!} = 3,360$

The word is "statistics". We fix two of the 's' letters at the beginning and end: s _ _ _ _ _ _ _ _ s.
This leaves 8 letters to arrange in the middle. The remaining letters are: t, a, t, i, s, t, i, c.
In this middle section of 8 letters, we have repetitions:
- 't' appears 3 times
- 'i' appears 2 times
- 's', 'a', 'c' appear 1 time each.
The number of distinct permutations for these 8 middle letters is calculated as:
$\frac{8!}{3!2!} = \frac{40,320}{(6)(2)} = \frac{40,320}{12} = 3,360$.

The positions on the advisory board are not distinct, so order does not matter. This means you should use combinations. For part (c), you will need to calculate the number of ways to choose from each group separately and then multiply the results together.

1. 21
2. 6
3. 12

1. We are choosing a group of 2 people from a total of 7 candidates. Since the order of selection doesn't matter, we use a combination: $\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{5040}{(2)(120)} = \frac{5040}{240} = 21$.
2. We are choosing 2 people from the 4 available biostatisticians. Again, order doesn't matter: $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{24}{(2)(2)} = 6$.
3. We must choose 1 epidemiologist from 3 AND 1 biostatistician from 4. We calculate the combinations for each and multiply them: $\binom{3}{1} \times \binom{4}{1} = \frac{3!}{1!2!} \times \frac{4!}{1!3!} = 3 \times 4 = 12$.

For Permutations: Think of this as arranging the 7 grades (A, B, B, B, C, C, F) in a line. This is a permutation with repeated items.
For Combinations: Think of this as a sequence of choices. First, choose which of the 7 students gets the A. Then, from the remaining students, choose which ones get the B's, and so on.

1. Permutation with repetitions: $\frac{7!}{1!3!2!1!} = 420$
2. Sequential combinations: $\binom{7}{1}\binom{6}{3}\binom{3}{2}\binom{1}{1} = 420$

1. Using Permutations: We are arranging 7 items (the grades). However, some grades are identical. This is a permutation with repetitions. The formula is $\frac{n!}{n_1!n_2!...n_k!}$, where $n$ is the total number of items, and $n_1, n_2, ...$ are the counts of each repeated item.
   Here, $n=7$, $n_A=1$, $n_B=3$, $n_C=2$, $n_F=1$.
   $\frac{7!}{1!3!2!1!} = \frac{5040}{1 \cdot 6 \cdot 2 \cdot 1} = \frac{5040}{12} = 420$.
2. Using Combinations: We can think of this as a sequence of tasks.
   1. Choose 1 student out of 7 to receive the A: $\binom{7}{1} = 7$ ways.
   2. Choose 3 students out of the remaining 6 to receive B's: $\binom{6}{3} = \frac{6!}{3!3!} = 20$ ways.
   3. Choose 2 students out of the remaining 3 to receive C's: $\binom{3}{2} = 3$ ways.
   4. Choose 1 student out of the remaining 1 to receive the F: $\binom{1}{1} = 1$ way.
   By the multiplication principle, the total number of ways is $7 \times 20 \times 3 \times 1 = 420$.

For part (a), you are choosing a group of 5 from 20. Does the order in which they are chosen matter for the final sample? This will tell you whether to use a combination or permutation.

1. $\binom{20}{5} = 15,504$
2. A combination is used because the sample is a group of individuals, and the order in which they are selected does not change the composition of the sample.

1. We are selecting a group of 5 people from a total of 20. Since the order of selection does not matter for the final sample, we use a combination:
   $\binom{20}{5} = \frac{20!}{5!(20-5)!} = \frac{20!}{5!15!} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = 15,504$.
   There are 15,504 possible samples.
2. A combination is used because a "study sample" refers to the final group of individuals, not the sequence in which they were picked. A sample containing persons {A, B, C, D, E} is identical to a sample containing {E, D, C, B, A}. Since the order does not create a different outcome, combinations are the appropriate method.

For part (a), you need to choose 1 person from the 3 who are diseased AND 4 people from the 17 who are not. For part (e), remember that the probability of "at least one" is equal to 1 minus the probability of "none".

1. 7,140
2. $P(A) = \frac{7140}{15504} \approx 0.4605$
3. $P(B) = \frac{6840}{15504} \approx 0.4412$
4. $P(A \cup B) = P(A) + P(B) \approx 0.9017$
5. $P(C) = 1 - P(B) \approx 0.5588$

There are 20 people total: 3 diseased and 17 not diseased. The total number of possible 5-person samples is $\binom{20}{5} = 15,504$.

1. To get a sample with exactly one diseased person, we must choose 1 of the 3 diseased people AND 4 of the 17 non-diseased people.
   Ways = $\binom{3}{1} \times \binom{17}{4} = 3 \times \frac{17!}{4!13!} = 3 \times 2380 = 7,140$.
2. $P(A) = \frac{\text{Ways to get 1 diseased}}{\text{Total possible samples}} = \frac{7140}{15504} \approx 0.4605$.
3. Event B is a sample with zero diseased individuals. This means we must choose 0 of the 3 diseased people AND 5 of the 17 non-diseased people.
   Ways to get 0 diseased = $\binom{3}{0} \times \binom{17}{5} = 1 \times \frac{17!}{5!12!} = 1 \times 6188 = 6,840$.
   $P(B) = \frac{6840}{15504} \approx 0.4412$.
4. This event is the union of event A (one diseased) and event B (zero diseased). In set notation, this is $A \cup B$. Since A and B are mutually exclusive (a sample can't have exactly one and exactly zero diseased people at the same time), the probability is $P(A \cup B) = P(A) + P(B) = \frac{7140}{15504} + \frac{6840}{15504} = \frac{13980}{15504} \approx 0.9017$.
5. Event C, "at least one diseased", is the complement of event B, "no diseased individuals". The easiest way to compute this is $P(C) = 1 - P(B)$.
   $P(C) = 1 - 0.4412 = 0.5588$. (Alternatively, one could calculate the probabilities of getting 1, 2, or 3 diseased individuals and add them up).