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3.7: Permutations and Combinations
Permutations In this section, we will develop an even faster way to solve some of the problems we have already learned to solve by other means. Let’s start with a couple examples.
Example 25
In this example, we needed to calculate n · ( n – 1) · ( n – 2) ··· 3 · 2 · 1. This calculation shows up often in mathematics, and is called the factorial , and is notated n!
Example 26
Now we will consider some slightly different examples.
Example 27
Factorial A factorial is when we take a positive integer and find the product of all descending positive integers, including itself, all the way to 1:
n! = n · ( n – 1) · ( n – 2) ··· 3 · 2 · 1
We say n! as “n factorial.” E.g., given 5!, then we say “5 factorial.”
How many different ways can the letters of the word MATH be rearranged to form a four- letter code word?
This problem is a bit different. Instead of choosing one item from each of several different categories, we are repeatedly choosing items from the same category (the category is: the letters of the word MATH) and each time we choose an item we do not replace it, so there is one fewer choice at the next stage: we have 4 choices for the first letter (say we choose A), then 3 choices for the second (M, T and H; say we choose H), then 2 choices for the next letter (M and T; say we choose M) and only one choice at the last stage (T). Thus, there are 4 · 3 · 2 · 1 = 24 ways to spell a code worth with the letters MATH.
How many ways can five different door prizes be distributed among five people?
There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be 5! = 5 · 4 · 3 · 2 · 1 = 120 ways.
A charity benefit is attended by 25 people and three gift certificates are given away as door prizes: one gift certificate is in the amount of $100, the second is worth $25 and the third is worth $10. Assuming that no person receives more than one prize, how many different ways can the three gift certificates be awarded?
Using the Basic Counting Rule, there are 25 choices for the person who receives the $ certificate, 24 remaining choices for the $25 certificate and 23 choices for the $10 certificate, so there are 25 · 24 · 23 = 13,800 ways in which the prizes can be awarded.
Example 28 Eight sprinters have made it to the Olympic finals in the 100-meter race. In how many different ways can the gold, silver and bronze medals be awarded?
Using the Basic Counting Rule, there are 8 choices for the gold medal winner, 7 remaining choices for the silver, and 6 for the bronze, so there are 8 · 7 · 6 = 336 ways the three medals can be awarded to the 8 runners.
Note that in these preceding examples, the gift certificates and the Olympic medals were awarded without replacement ; that is, once we have chosen a winner of the first door prize or the gold medal, they are not eligible for the other prizes. Thus, at each succeeding stage of the solution there is one fewer choice (25, then 24, then 23 in the first example; 8, then 7, then 6 in the second). Contrast this with the situation of a multiple-choice test, where there might be five possible answers — A, B, C, D or E — for each question on the test.
Note also that the order of selection was important in each example: for the three door prizes, being chosen first means that you receive substantially more money; in the Olympics example, coming in first means that you get the gold medal instead of the silver or bronze. In each case, if we had chosen the same three people in a different order there might have been a different person who received the $100 prize, or a different gold medalist. (Contrast this with the situation where we might draw three names out of a hat to each receive a $10 gift certificate; in this case, the order of selection is not important since each of the three people receive the same prize. Situations where the order is not important will be discussed in the next section.)
We can generalize the situation in the two examples above to any problem without replacement where the order of selection is important. If we are arranging in order r items out of n possibilities (instead of 3 out of 25 or 3 out of 8 as in the previous examples), the number of possible arrangements will be given by
n · ( n – 1) · ( n – 2) ··· ( n – r + 1)
If you don’t see why ( n — r + 1) is the right number to use for the last factor, just think back to the first example in this section, where we calculated 25 · 24 · 23 to get 13,800. In this case n = 25 and r = 3, so n — r + 1 = 25 — 3 + 1 = 23, which is exactly the right number for the final factor.
Now, why would we want to use this complicated formula when it’s actually easier to use the Basic Counting Rule, as we did in the first two examples? Well, we won’t actually use this formula all that often, we only developed it so that we could attach a special notation and a special definition to this situation where we are choosing r items out of n possibilities without replacement and where the order of selection is important.
P
Using the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are 25 · 24 · 23 = 13,800 ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn’t really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6:
ABC ACB BAC BCA CAB CBA
How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are 3 · 2 · 1 = 6 ways to do this; we didn’t really need to list them all, we can just use permutations!
So, out of the 13,800 ways to select 3 people out of 25, six of them involve Abe, Bea and Cindy. The same argument works for any other group of three people (say Abe, Bea and David or Frank, Gloria and Hildy) so each three-person group is counted six times. Thus the 13,800 figure is six times too big. The number of distinct three-person groups will be 13,800/6 = 2300.
We can generalize the situation in this example above to any problem of choosing a collection of items without replacement where the order of selection is not important. If we are choosing r items out of n possibilities (instead of 3 out of 25 as in the previous
examples), the number of possible choices will be given by
n Pr (^) , and we could use this r Pr formula for computation. However, this situation arises so frequently that we attach a special notation and a special definition to this situation where we are choosing r items out of n possibilities without replacement where the order of selection is not important.
Combinations C = n^ Pr n r r r
We say that there are (^) n C (^) r combinations of size r that may be selected from among n choices without replacement where order doesn’t matter.
We can also write the combinations formula in terms of factorials:
n C^ r =^
n! ( n − r )! r!
Example 32 A group of four students is to be chosen from a 35-member class to represent the class on the student council. How many ways can this be done?
Since we are choosing 4 people out of 35 without replacement where the order of selection is
not important there are 35 C (^4) = 35 ⋅^34 ⋅^33 ⋅^32 4 ⋅ 3 ⋅ 2 ⋅ 1
= 52,360 combinations.
Try it Now 2 The United States Senate Appropriations Committee consists of 29 members; the Defense Subcommittee of the Appropriations Committee consists of 19 members. Disregarding party affiliation or any special seats on the Subcommittee, how many different 19-member subcommittees may be chosen from among the 29 Senators on the Appropriations Committee?
In the preceding Try it Now problem we assumed that the 19 members of the Defense Subcommittee were chosen without regard to party affiliation. In reality this would never happen: if Republicans are in the majority they would never let a majority of Democrats sit on (and thus control) any subcommittee. (The same of course would be true if the Democrats were in control.) So, let’s consider the example again, in a slightly more complicated form.
Example 33 The United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee?
In this case, we need to choose 10 of the 15 Republicans and 9 of the 14 Democrats. There are 15 C 10 = 3003 ways to choose the 10 Republicans and 14 C 9 = 2002 ways to choose the 9 Democrats. But now what? How do we finish the problem?
Suppose we listed all of the possible 10-member Republican groups on 3003 slips of red paper and all of the possible 9-member Democratic groups on 2002 slips of blue paper. How many ways can we choose one red slip and one blue slip? This is a job for the Basic Counting Rule! We are simply making one choice from the first category and one choice from the second category, just like in the restaurant menu problems from earlier.
There must be 3003 · 2002 = 6,012,006 possible ways of selecting the members of the Defense Subcommittee.
As above, the number of possible outcomes of the lottery drawing is 48 C 6 = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by 6 C 5 = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by 42 C 1 = 42. Thus, the number of favorable outcomes is then given by the Basic Counting Rule: 6 C 5 · 42 C 1 = 6 · 42 = 252. So, the probability of winning the second prize is.
( 6 C 5 )( 42 C 1 )
48 C 6 12271512
This means there is less than a 1% chance of winning second prize. Wow! We now can see why some people call it “luck” when winning the lottery because the chances of winning are so low.
Try it Now 3 A multiple-choice question on an economics quiz contains 10 questions with five possible answers each. Compute the probability of randomly guessing the answers and getting 9 questions correct.
Example 37
Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.
In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus, we use combinations to compute the possible number of 5-card hands, 52 C 5. This number will go in the denominator of our probability formula, since it is the number of possible outcomes.
For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Since there are four Aces and we want exactly one of them, there will be 4 C 1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48 C 4 ways to select the four non-Aces. Now we use the Basic Counting Rule to calculate that there will be 4 C 1 · 48 C 4 ways to choose one ace and four non-Aces.
Putting this all together, we have
P (one Ace) = (^4 C^1 )(^48 C^4 )^ = 778320 ≈ 0.
52 C 5 2598960
Thus, there is about a 30% chance of drawing exactly one Ace in a hand of 5 cards.
Example 38 Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.
The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:
P (two Aces) = (^4 C^2 )(^48 C 3 )^ = 103776 ≈ 0. 52 C 5 2598960
There is about a 4% chance of drawing 2 Aces in a hand of 5 cards. Notice from example 37, the probability of drawing exactly one Ace is much higher that drawing two.
It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.
Try it Now 4 Compute the probability of randomly drawing five cards from a deck of cards and getting three Aces and two Kings.
Birthday Problem Let’s take a pause to consider a famous problem in probability theory:
Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?
Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let’s see if we should listen to our intuition. Let’s start with a simpler problem, however.
Example 39 Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people?
There are a lot of ways there could be at least one shared birthday. Fortunately, there is an easier way. We ask ourselves “What is the alternative to having at least one shared birthday?” In this case, the alternative is that there are no shared birthdays. In other words, the alternative to “at least one” is having none. In other words, since this is a complementary event,
P(at least one) = 1 – P(none)
We will start, then, by computing the probability that there is no shared birthday. Let’s imagine that you are one of these three people. Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that
Example 41
If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn’t studied probability!) You wouldn’t be guaranteed to win, but you should win more than half the time.
This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts. If you still don’t believe the math, you can carry out a simulation. Just so you won’t have to go around rounding up groups of 30 people, someone has kindly developed a Java applet so that you can conduct a computer simulation. Go to this web page: http://www-stat.stanford.edu/~susan/surprise/Birthday.html, and once the applet has loaded, select 30 birthdays and then keep clicking Start and Reset. If you keep track of the number of times that there is a repeated birthday, you should get a repeated birthday about 7 out of every 10 times you run the simulation.
Try it Now 5 Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?
Suppose 30 people are in a room. What is the probability that there is at least one shared birthday among these 30 people?
Here we can calculate
P (shared birthday) = 1 − 365
P 30
which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday.
Try it Now Answers
- There are 26 characters. a. 26^5 = 11,881,376. b. 26 P 5 = 26·25·24·23·22 = 7,893,
- Order does not matter. 29 C 19 = 20,030,010 possible subcommittees
- (^) There are 5^10 = 9,765,625 different ways the exam can be answered. There are 9 possible locations for the one missed question, and in each of those locations there are 4 wrong answers, so there are 36 ways the test could be answered with one wrong answer. 36 P(9 answers correct) = 5 10
≈ 0.0000037 chance
4. P (three Aces and two Kings ) = (^4 C^3 )( 4 C 2 )^
52 C 5 2598960
- (^) P (shared birthday ) = 1 − 365
P 10
Exercises
- A boy owns 2 pairs of pants, 3 shirts, 8 ties, and 2 jackets. How many different outfits can he wear to school if he must wear one of each item?
- At a restaurant you can choose from 3 appetizers, 8 entrees, and 2 desserts. How many different three-course meals can you have?
- How many three-letter words can be made from 4 letters FGHI if a. repetition of letters is allowed b. repetition of letters is not allowed
- How many four-letter words can be made from 6 letters AEBWDP if a. repetition of letters is allowed b. repetition of letters is not allowed
- All of the license plates in a particular state feature three letters followed by three digits (e.g. ABC 123). How many different license plate numbers are available to the state’s Department of Motor Vehicles?
- A computer password must be eight characters long. How many passwords are possible if only the 26 letters of the alphabet are allowed?
- You own 16 CDs. You want to randomly arrange 5 of them in a CD rack. What is the probability that the rack ends up in alphabetical order?
- A jury pool consists of 27 people, 14 men and 13 women. Compute the probability that a randomly selected jury of 12 people is all male.
- In a lottery game, a player picks six numbers from 1 to 48. If 5 of the 6 numbers match those drawn, they player wins second prize. What is the probability of winning this prize?
- In a lottery game, a player picks six numbers from 1 to 48. If 4 of the 6 numbers match those drawn, they player wins third prize. What is the probability of winning this prize?
- Compute the probability that a 5-card poker hand is dealt to you that contains all hearts.
- Compute the probability that a 5-card poker hand is dealt to you that contains four Aces.
*Adapted from Math for Liberal Arts Students by David Lipman & Darlene Diaz
SOLUTIONS
1. 96 2. 48 3. 64, 4. 1296, 360 5. 17576000 6. 208827064576 7. 24 8. 9129120 9. 840 10. ~1.19x10^13 11. 95040 12. 479001600 13. 495 14. 6188 15. 15890700 16. 120 17. 208606320 18. 163350 19. 0. 20. 5.2x10 - 21. 2.054x10- 22. 0. 23. 4.95x10- 24. 1.85x10 -
