Permutations and combinations

This is the second post in a series of posts on combinatorial analysis. The first post is on the multiplication principle. This post introduces permutations and combinations.

The starting point of the discussion on counting is the idea that if one event can occur in M ways and another event can occur, independent of the first event, in N ways, then the two events together can occur in M \times N ways. This idea is often called the multiplication principle and is introduced in this previous post.

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Factorial

Consider the eight letters A, B, C, D, E, F, G and H. How many ways can we order the 8 letters? The following shows two ordered arrangements.

    A-B-C-D-E-F-G-H

    C-B-F-A-D-G-H-E

The multiplication principle can help us determine the number of all arrangements of 8 letters. We are trying to fill 8 objects into 8 positions. There are 8 choices for the first position. After the first letter is fixed, there are 7 letters for the second position. Continue with this approach, there will be only one letter left for the eighth position. In total, there are 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320 ordered arrangements of 8 letters. A convenient notation for the answer is 8!, read 8 factorial. It is simply the product of the integers from 8 down to 1.

In general, if n is a positive integer, n!, read n factorial, is the product of all the integers from n down to 1.

n!=n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1

For the above definition of n! to make sense, n is a positive integer. We define 0! = 1 to make certain formulas easy to calculate.

According to the above example, there are 8! many different ordered arrangements of 8 distinct objects (objects that are distinguishable from one another). Any ordered arrangement such as C-B-F-A-D-G-H-E is called a permutation of the 8 letters. In general, a permutation is an ordered arrangement of a set of objects that are distinguishable from one another. From the above example of 8 letters, we have the following observation.

    The total number of permutations of n distinct objects is n!.

The G8 summit is an annual meeting between leaders from eight of the most powerful countries in the world. There would be 8! = 40,320 different ways to line up the 8 leaders for a group photo. A permutation can be viewed as ranking. In a soccer league of 8 teams, there would be 8! = 40,320 different possible rankings.

The factorial grows very rapidly. In ordering 9 objects, there are 9! = 362,880 permutations.

    10! = 3,628,800
    11! = 39,916,800
    12! = 479,001,600
    15! = 1,307,674,368,000
    17! = 355,687,428,096,000
    20! = 2,432,902,008,176,640,000

Thus the number of possible rankings of 12 job applicants is over 479 millions. The number of possible rankings of 20 job applicants would be a staggering number, over 2.4 quintillions! A quintillion is 10 raised to 18. In contrast, a trillion is 10 raised to 12.

Example 1
Suppose that you have 12 books with 4 of them being novels, 4 of them being math books, 3 of them being science books and one being a history book. You want to place the books on the same level in a bookshelf. If the books are put in any order, then there would be 12! = 479,001,600 different placements of books. In how many ways can the books be placed in the bookshelf if all books of the same subject are to be together?

The last question makes use of the multiplication principle in a subtle way. Consider the books arranged by subjects in this order: novel-math-science-history. By the multiplication principle, there are 4! x 4! x 3! x 1! = 24 x 24 x 6 x 1 = 3,456 many ordered arrangements. This is just for one specific way of the order of the 4 subjects – novel-math-science-history. There are 4! = 24 ways of order 4 subjects. Then the answer to the last question is 24 x 3,456 = 82,944. \square

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Permutation

Sometimes we do not order the entire set of n objects. We only order a subset of the n objects. Using the above example of 8 letters A, B, C, D, E, F, G and H, in how many ways can we arrange three letters from these 8 letters? The multiplication principle is at work here. We are attempting to fill 3 positions using 8 letters at our disposal. The first position has 8 choices. The second position has 7 choices and the third position has 6 choices. There are a total of 8 x 7 x 6 = 336 ordered arrangements. Three examples are A-B-C, G-A-H and E-B-F. Any one of these 336 ordered arranged is called 3-permutation of 8 letters. We can also call it a permutation of 8 letters taken 3 at a time.

Example 2
To put the 8 letters in a context, suppose that a math club has 8 students – Abby, Beth, Chloe, Diane, Edward, Frank, George and Henry. In how many ways can we choose 3 students to fill three positions of President, Vice President and Treasurer of the math club? Based on the preceding discussion, the answer is 8 x 7 x 6 = 336. The following is a presentation of the calculation that gives the hint for a formula.

    \displaystyle 8 \times 7 \times 6=\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1}=\frac{8!}{5!}=\frac{8!}{(8-3)!}

In general, any ordered arrangement of k objects from a set of n objects is called a k-permutation of n objects. We can also call any such ordered arrangement a permutation of n objects taken k at a time. Here, we attempt to fill k positions with objects selected from n objects with k \le n. How many permutations of n objects taken k at a time? There are

    n \times (n-1) \times (n-2) \times \cdots \times (n-(k-1))

many different permutations of n objects taken k at a time. Some common notations for this number are P(n,k) or P_{n,k}. According to the hint in Example 2, the following is a formula for P(n,k).

\displaystyle P(n,k)=\frac{n!}{(n-k)!} is the number of permutations of n objects taken k at a time.

Note that P(n,n) is simply n!.

Example 3
In evaluating 20 job applicants, there are 20! many possible rankings, which is over 2 quintillions as discussed above. A more sensible goal is to narrow the list to the top 5 candidates. The number of permutations of 20 candidates taken 5 at a time is

    \displaystyle P(20,5)=\frac{20!}{15!}=20 \times 19 \times 18 \times 17 \times 16=\text{1,860,480}

Example 4
Suppose that the psychology club at a university has 12 members in the lower division (consisting of freshmen and sophomores), 10 members in the upper division (consisting of juniors and seniors). Four awards (achievements in terms of overall grade, leadership potential, volunteerism, and public speaking) are given to each group. What is the total number of different outcomes if

  • a student can receive more than one award,
  • a student can receive only one award.

If a student can be awarded more than once, then there are 12^4 ways of giving out awards in the lower division and 10^4 ways in the upper division, leading to the following answer.

    \displaystyle 12^4 \times 10^4 = \text{207,360,000}

If a student can only be awarded once, then the following is the answer.

    \displaystyle P(12,4) \times P(10,4) =(12 \times 11 \times 10 \times 9) \times (10 \times 9 \times 8 \times 7)= \text{59,875,200}

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When Some Objects are Indistinguishable

In calculating the number of permutations, the objects are assumed to be distinct. We now look at the situation where objects are separated into groups of indistinguishable objects. The goal is still finding the number of ordered arrangements. The number of permutations as calculated above would be an over count. The answer is obtained by dividing the inflated count by the amount of over counting.

Example 5
How many different letter arrangements can be formed from the letters in the word BANANA?

There are 6 letters in BANANA. Three of the letters are indistinguishable (A) and two of them are indistinguishable (N). If all the letters are distinguishable, then there are 6! permutations. But 6! = 720 over counts. By how many?

Pretend all the letters are distinguishable. Take the arrangement A_1 \ A_2 \ A_3 \ B \ N_1 \ N_2. Now permute the A’s and the N’s. Doing so should result in 3! x 2! = 12 permutations.

    A_1 \ A_2 \ A_3 \ B \ N_1 \ N_2
    A_1 \ A_3 \ A_2 \ B \ N_1 \ N_2
    A_2 \ A_1 \ A_3 \ B \ N_1 \ N_2
    A_2 \ A_3 \ A_1 \ B \ N_1 \ N_2
    A_3 \ A_1 \ A_2 \ B \ N_1 \ N_2
    A_3 \ A_2 \ A_1 \ B \ N_1 \ N_2

    A_1 \ A_2 \ A_3 \ B \ N_2 \ N_1
    A_1 \ A_3 \ A_2 \ B \ N_2 \ N_1
    A_2 \ A_1 \ A_3 \ B \ N_2 \ N_1
    A_2 \ A_3 \ A_1 \ B \ N_2 \ N_1
    A_3 \ A_1 \ A_2 \ B \ N_2 \ N_1
    A_3 \ A_2 \ A_1 \ B \ N_2 \ N_1

But in reality, all of the 12 permutations are the same: A \ A \ A \ B \ N \ N. So each unique permutation appears 12 times in the count 6!. Thus the correct count is

    \displaystyle \frac{6!}{3! \ 2! \ 1!}=60

In general, \displaystyle \frac{n!}{m_1! \ m_2! \ \cdots \ m_k!} is the number of permutations of n objects, of which m_1 are alike, m_2 are alike, …, m_k are alike, where m_1+\cdots+m_k=n.

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Combination

In all the above instances of counting, the focus is to find all ordered arrangements (or permutations) of a set of objects. We now focus on the different subsets of k elements that can be chosen from a set of n objects. In this setting, the goal is not to count ordered arrangements. The goal is to count the number of subsets of a certain size. For example, let’s say you want to buy a 2-scoop ice cream cone of different flavors chosen from 3 flavors – Chocolate (C), Strawberry (S) and Vanilla (V). Suppose you do not care which flavor is on top and which one is at the bottom. The different subsets of size 2 would be: CS, CV, and SV. Here, CS is not an ordered arrangement. It just means that the two chosen flavors are chocolate and strawberry. If order matters, then there would be six outcomes – CS, SC, CV, VC, SV, and VS. So the approach is to find the number of permutations (an over count) and then divide the amount of over count.

Example 6
Suppose that a math club has 8 students – Abby, Beth, Chloe, Diane, Edward, Frank, George and Henry. In how many ways can we choose three students to form a committee of three?

The number of permutations of three students is 8 x 7 x 6 = 336. This count is inflated since there are many repeats reflected in this count. For example, if the chosen students are Abby, Beth and Chloe, then there are 3! = 6 ways to order these three students – ABC, ACB, BAC, BCA, CAB and CBA. So each set of three students is reflected 6 times in the inflated count of 336. So the answer is 336 / 6 = 56. Thus there are 56 possible groups of three students when three students are chosen from 8 students. The following calculation makes the formula clear.

    \displaystyle \frac{8 \times 7 \times 6}{3!}=\frac{8 \times 7 \times 6 \times 5!}{3! \ 5!}=\frac{8!}{3! \ 5!}

In general, \displaystyle \binom{n}{k}=\frac{n!}{k! \ (n-k)!} is the number of different groups of size k that could be chosen from a set of n distinct objects where k=0,1,2,\cdots,n.

The number \binom{n}{k} is called a binomial coefficient and is pronounced “n choose k.” Other notations for \binom{n}{k} are _nC_k, C_{n,k} and C(n,k).

Example 7
A committee of five is to be chosen from a group of 11 people consisting of 5 men and 6 women.

  • How many different committees of five people can be formed?
  • How many different committees consisting of 2 men and 3 women can be formed?
  • How many different committees of five consisting of either 2 men or 2 women?

The number of possible committees of five would be \binom{11}{5}=462. For the second problem, note that there are \binom{5}{2}=10 possible groups of 2 men and there are \binom{6}{3}=20 groups of 3 women. By the multiplication principle, there are 10 x 20 = 200 different groups of 5 people consisting of 2 men and 3 women.

For the third problem, there are two cases: a committee of 2 men and 3 women or a committee of 2 women and 3 men. The answer is the sum of these two cases. Thus the answer is

    \displaystyle \binom{5}{2} \times \binom{6}{3}+\binom{5}{3} \times \binom{6}{2}=10 \times 20+10 \times 15=350

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More Examples

We present a few more examples to demonstrate the ideas discussed here. Exercises are given at the end.

Example 8
A child plays with a set of 12 blocks, 4 of which are red, 4 of which are black, 3 of which are white and one of which is yellow. In how many ways can these blocks be arranged in a row? In how many ways can these blocks be arranged in a row in such a way that all blocks of the same color are together?

Note the similarity with Example 1. Both problems concern with a collection of 12 objects that can be separated into 4 sub groups. In Example 1, the objects in each sub group are distinguishable while the objects in this example are indistinguishable in each sub group. As a result, the formula discussed in the section “When Some Objects are Indistinguishable” would apply. Here’s the number of different ways to order the blocks.

    \displaystyle \frac{12!}{4! \ 4! \ 3! \ 1!}=\text{138,600}

For the second question, the answer would be 4! = 24. Unlike Example 1, we do not need to order the blocks within the same color since they are indistinguishable. \square

Example 9
The tennis team in School A has 12 players and the tennis team in School B has 14 players. If 5 players are selected from each team and then paired off for a tennis game, how many results are possible?

There are \binom{12}{5}=792 ways of choosing 5 players from Team A. There are \binom{14}{5}=2002 ways of choosing 5 players from Team B. There are 5! = 120 ways in pairing off the two teams of 5 players. Thus there are 792 x 2002 x 120 = 190,270,080 ways of choosing two teams and then pairing them off for a tennis game. \square

Example 10
A company is made up of 3 departments with the one department consisting of 8 employees, the second department consisting of 10 employees and the third department consisting of 5 employees. The owner of the company decides to promote two employees chosen from the 23 employees. Suppose that the two employees are selected at random. How many ways can the two employees be selected if

  • both promotions are from the same department?
  • both promotions are from different departments?

The problem does not mention the positions for the promotion. Thus we assume that the two positions are identical. The first question has three cases based on the departments. The following is the sum of the three cases.

    \displaystyle \binom{8}{2}+\binom{10}{2}+\binom{5}{2}=28+45+10=83

For the second question, there are \binom{3}{2}=3 cases, based on pairs of departments. The following is the sum of the 3 cases.

    \displaystyle \binom{8}{1} \times \binom{10}{1}+\binom{8}{1} \times \binom{5}{1}+\binom{10}{1} \times \binom{5}{1}=80+40+50=170

Example 11
The problem is the same as Example 10 except that the two promotions are for President and Vice President.

Now that the two promotions are different, order matters. Thus use permutations to perform the calculations. the answer to the first question is:

    \displaystyle \frac{8!}{(8-2)!}+\frac{10!}{(1-2)!}+\frac{5!}{(5-2)!}=56+90+20=166

In the second question, we still pick one person from each department, but it needs to be multiplied by two since the positions are different.

    \displaystyle 2 \times 8 \times 10+2 \times 8 \times 5+2 \times 10 \times 5=160+80+100=340

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Exercises

Exercise 1
How many different letter arrangements can be formed from the letters in the word MISSISSIPPI?

Exercise 2
In how many ways can a manager assigns 3 different jobs at random to 11 available workers?

Exercise 3
Suppose that the license plates for a certain state consist of three numbers and three letters. How many different license plates are possible if

  1. the letters must come before the numbers?
  2. all letters must appear together and all numbers must appear together?
  3. only the letters must appear together?
  4. there is no restriction on the positions of the letters and numbers?

Exercise 4
In ordering a pizza, cheese and tomato sauce come with the pizza. The customer can also choose to add toppings from the following lists:

    Meat items: ham, sausage, pepperoni, salami, chicken.
    Vegetable items: pineapple, olive, mushroom, onion, bell peppers.

  1. How many distinct pizzas are possible?
  2. Suppose that a customer does not like salami and sausage in the meat list and onion and olive in the vegetable list and thus will not add these toppings. How many different pizzas are possible?
  3. Suppose that a customer would like to add two meat items and three vegetable items as toppings. How many different pizzas are possible?
  4. Suppose that a customer would like to add exactly 5 toppings. How many different pizzas are possible?
  5. Suppose that a customer would like to add at most 5 toppings. How many different pizzas are possible?

Exercise 5

  1. Twelve workers are randomly assigned to work in four factories with 3 workers assigned to each factory as assembly line workers. How many ways can the workers be assigned to the factories?
  2. Twelve workers are randomly assigned to work in four factories with 3 workers assigned to each factory as manager, assistant manager and assembly line worker. How many ways can the workers be assigned to the factories?

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Answers

Exercise 1

    34,650

Exercise 2

    990

Exercise 3

  1. 17,576,000
  2. 35,152,000
  3. 70,304,000
  4. 351,520,000

Exercise 4

  1. 1024
  2. 64
  3. 100
  4. 252
  5. 638

Exercise 5

  1. 369,600
  2. 479,001,600

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\copyright \ 2017 \text{ by Dan Ma}

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The multiplication principle

This is the first post in a series of posts on combinatorial analysis, which is a study on counting, e.g. finding effective methods for counting the number of ways to arrange objects and for counting the number of ways events can occur. Many problems in mathematics, e.g. in probability theory, require the counting of the number of ways an event can occur. Permutations and combinations are basic ideas in counting. Both ideas are based on an idea called the multiplication principle. This post focuses on the multiplication principle. The next post is on permutations and combinations.

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The Multiplication Principle

Let’s start with some examples.

Example 1
Jack is buying ice cream – two scoops of ice cream of the same flavor on either a cone or a cup. The available flavors are vanilla, strawberry, chocolate, butter pecan, mint chocolate chips and rocky road. How many possible arrangements of ice cream are available for Jack?

The answer is 2 x 6 = 12 arrangements. The following lists out all the possibilities.

    Cone-Vanilla
    Cone-Strawberry
    Cone-Chocolate
    Cone-Butter Pecan
    Cone-Mint Chocolate Chips
    Cone-Rocky Road

    Cup-Vanilla
    Cup-Strawberry
    Cup-Chocolate
    Cup-Butter Pecan
    Cup-Mint Chocolate Chips
    Cup-Rocky Road

With the cone option, there are 6 choices for flavors. With the cup option, there are also 6 choices for flavors. So the total number of possible choices would be 6 + 6 = 2 x 6 = 12.

Of course, if Jack only limits his purchase to his most favorite ice cream (say mint chocolate chips), then there would be only two choices, cone or cup. But if he is indifferent to the flavors, there would be 12 arrangements as listed above. \square

The listing in Example 1 suggests a general principle.

    The Multiplication Principle
    If one event can occur in M ways and another event can occur independent of the first event in N ways, then the two events successively can occur in M \times N ways.

Of course, the principle just described has an obvious generalization. Instead of just considering two events, we can count how many ways a series of events can occur. As long as the events are independent, the total number of ways all events occurring in a row is obtained by simply multiplying the numbers of possible occurrences of the individual events.

Example 2
The following is a menu for a banquet. Assuming that each guest can only pick one item for each course, how many different possible dinner decisions are possible?

    menu-for-counting

This problem is mentioned in previous post. Using the multiplication principle, the answer is 4 x 2 x 3 x 2 = 48 choices. \square

Remark. For the multiplication principle to apply, the events must be independent. This means that the number of occurrences of the one event is not restricted by the previous event. If a guest in the banquet is indifferent to all items in the menu (if taken individually) but does not like to mix certain items (e.g. he does not like to have French onion soup to go with goat cheese salad), then he or she might not have 48 choices. The following example illustrates that for a certain kind of dependency between events, the multiplication principle can still apply.

Example 3
A PIN code for a bank ATM machine is a 4-digit number. Determine the following:

  • How many PIN codes are possible?
  • How many PIN codes are possible if no repetition of digits is allowed?
  • How many PIN codes are possible if the first two digits form a number less than or equal to 33 and the last two digits form a number that is at least 3 times the first two digits?

There are 10 \times 10 \times 10 \times 10= 10^4=10,000 many possible codes. If no digits can repeat, then there are

    10 \times 9 \times 8 \times 7=5040

different codes. For the no repetition case, there are 10 choices for the first digit. Once the first digit is fixed, there are only 9 choices for the second digit. Once the first two digits are fixed, there are 8 choices for the third digit. Then the fourth digit has 7 choices once the first three digits are fixed. Though the number of choices for each digit is limited by the previous digit, it is the same regardless the value of the previous digits. Thus the principle still applies.

The third problem illustrates a dependency that makes the multiplication principle not applicable. For example, if the first two digits are 01, the last two digits would be 03, 04, 05, 06, …, 99 for a total of 97 possibilities. If the first two digits are 02, then there are 94 possibilities for the last two digits (06, 07, …, 99). As the number for the first two digits increases to 33, the number of choices for the last two digits would decrease. The largest possibility for the first two digits would be 33 and there is only one choice for the last two digits (99). Thus the multiplication principle would not work here. The answer is not obtained by multiplication. It is obtained by adding up all the possibilities as the first two digits go from 01 to 33. We perform the calculation in Excel and the answer is 1617. Of course, the third problem is not a realistic way to obtain pass codes. It is just an illustration of a case where the multiplication principle does not apply. \square

We look at more examples.

Example 4
In how many ways can 9 students (4 males and 5 females) stand in a row for a group photo

  • if there is no restriction on the standing positions?
  • if the females must stand together and the males must stand together?
  • if the females must stand together?

If the students can stand anywhere they want, there are 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362,880 ways of arranging these 9 students in a row. For the first spot, there are 9 students to choose from and there are 8 students to choose from for the second spot and so on. A convenient notation for the answer is 9! (read 9 factorial). It simply means that it is the product of all the integers from 9 down to 1.

By the same reasoning, there are 5! = 120 ways of arranging 5 persons in a row. There are 4! = 24 ways of arranging 4 people standing in a row. If the females stand on the left and the males stand on the right, then there are 120 x 24 = 2,880 arrangements in a row for the 9 students. The answer is 2 x 120 x 24 = 5,760 since there is also the case of males standing on the left and females on the right.

The third problem requires that only the females must stand together. The use of the multiplication principle is subtle. If the females must stand together, think of all the females as one unit. There are 5 units – 4 males and one combined unit of females. There are 5! ways of arranging the 5 units. Within the combined unit of females, there are 5! ways of arranging the 5 females. Thus the answer is 5! x 5! = 120 x 120 = 14,400. \square

The next example shows that the multiplication principle is invaluable in calculating large numbers.

Example 5
In general, longer passwords are stronger passwords. The principle of counting discussed here can be used to quantify the strength. Consider an 8-character password consisting of letters from English. To start, let’s say the letters in the password are all upper case or all lower case, i.e. the password is case insensitive. Then the number of possible passwords would be

    \displaystyle 26 \times 26 \times 26 \times 26 \times 26 \times 26 \times 26 \times 26=26^8=\text{208,827,064,576}

That’s over 208.8 billions possibilities. What if the password is case sensitive? If each character in the password can be in upper case or in lower case, then each character has 52 choices (26 upper case letters and 26 lower case letters). Then the number of possible passwords would be

    \displaystyle 52 \times 52 \times 52 \times 52 \times 52 \times 52 \times 52 \times 52=52^8=\text{53,459,728,531,456}

That’s over 53 trillions possible passwords! Of course, the length of the password by itself may not mean strength. For example, people may use familiar words to form the password. One of the famous examples would have to be “Password”. In addition to being long enough, the characters in the passwords should be randomly chosen if possible.

To make the 8-character password even stronger, we can include numbers in addition to letters. Then the total number of possibilities is

    \displaystyle 62^8=\text{218,340,105,584,896}

which is over 218 trillions. \square

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Factorial

The factorial concept introduced in Example 4 is a useful tool in counting. If n is a positive integer, the number n! is defined to be the product of all the positive integers less than or equal to n.

    n!=n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1

For convenience, we define 0! = 1. In Example 4, 9! is the number of ways of ordering 9 people in a row. In general, n! is the number of ordered arrangements of n distinct objects. There will be more discussion about this interpretation of n factorial in the next post.

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Exercises

Exercise 1
Example 3 and Example 5 concern with finding the number of possible ordered arrangements of objects for use in setting passwords or pass codes. Here’s is an example involving license plates. When a motor vehicle authority (or some relevant governmental body) designs license plates for automobiles, one important goal is to have a sufficiently large number of possibilities to accommodate all the vehicles in that jurisdiction. Suppose that the license plate has 7 characters. The first character is a number and the next three characters are letters. The last three characters are numbers.

  1. What is the total number of possibilities in this license plate design?
  2. If the motor vehicle authority is beginning to issue license plates with the first digit being 3, how many license plates will the authority have to issue before switching to using 4 as the first digit?

Exercise 2

  1. Twelve tasks are to be assigned to twelve workers. What is the total different number of job assignments?
  2. In how many ways can four students be randomly selected to receive four cash prizes ($10, $20, $30, $40)?

Exercise 3

  1. In how many ways can 4 boys and 4 girls stand together in a row for a group photo?
  2. In how many ways can 4 boys and 4 girls stand together in a row for a group photo if people of the same gender must be together?
  3. In how many ways can 4 boys and 4 girls stand together in a row for a group photo if only the girls must be together?
  4. In how many ways can 4 boys and 4 girls stand together in a row for a group photo if no people of the same gender can stand together?

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Answers

Exercise 1

  • 175,760,000
  • 17,576,000

Exercise 2

  • 479,001,600
  • 11,880

Exercise 3

  • 40,320
  • 1,152
  • 2,880
  • 1,152

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\copyright \ 2017 \text{ by Dan Ma}

A banquet menu can be a lesson in counting

    menu-for-counting

The above is a menu found in the Internet. Though we do not know how the dishes taste, the menu seems enticing. Whatever the eventual quality of the food, it is a well crafted menu. There are choices for each course. Assuming you can only pick one choice for each course, how many possible choices does a guest have for a complete meal in this banquet?

The above question is a counting problem, which in this case is easy to answer. The answer is derived by multiplying a few numbers. However, the significance of the problem is broader than the procedure to get the answer. The fundamental ideas behind this and other “easy” examples form a foundation of the mathematical subject called combinatorics. In order to help build a firm foundation, we devote several posts introducing the subject of counting, starting from the fundamental notions.

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\copyright \ 2017 \text{ by Dan Ma}