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Like the Combinations Calculator the Permutations Calculator finds the number of subsets that can be taken from a larger set. However, the order of the subset matters. The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders.

1. As with permutations, the calculator provided only considers the case of combinations without replacement, and the case of combinations with replacement will not be discussed. Using the example of a soccer team again, find the number of ways to choose 2 strikers from a team of 11. Unlike the case given in the permutation example, where the.
2. WinDrawWin.com - Free football betting calculators. Betting Calculators. Check out this comprehensive range of bet winnings calculators to help you with all those common betting permutations.

Examples: Probability using Permutations and Combinations. We can use permutations and combinations to help us answer more complex probability questions. 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. The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders.

Factorial

There are n! ways of arranging n distinct objects into an ordered sequence, permutations where n = r.

Combination

The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed.

Permutation

The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed. When n = r this reduces to n!, a simple factorial of n.

Combination Replacement

The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are allowed.

Permutation Replacement

The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are allowed.

n

the set or population

r

subset of n or sample set

Permutations Formula:

For n ≥ r ≥ 0.

Calculate the permutations for P(n,r) = n! / (n - r)!. 'The number of ways of obtaining an ordered subset of r elements from a set of n elements.'[1]

Permutation Problem 1

Choose 3 horses from group of 4 horses

In a race of 15 horses you beleive that you know the best 4 horses and that 3 of them will finish in the top spots: win, place and show (1st, 2nd and 3rd). So out of that set of 4 horses you want to pick the subset of 3 winners and the order in which they finish. How many different permutations are there for the top 3 from the 4 best horses?

For this problem we are looking for an ordered subset of 3 horses (r) from the set of 4 best horses (n). We are ignoring the other 11 horses in this race of 15 because they do not apply to our problem. We must calculate P(4,3) in order to find the total number of possible outcomes for the top 3 winners.

P(4,3) = 4! / (4 - 3)! = 24 Possible Race Results

If our 4 top horses have the numbers 1, 2, 3 and 4 our 24 potential permutations for the winning 3 are {1,2,3}, {1,3,2}, {1,2,4}, {1,4,2}, {1,3,4}, {1,4,3}, {2,1,3}, {2,3,1}, {2,1,4}, {2,4,1}, {2,3,4}, {2,4,3}, {3,1,2}, {3,2,1}, {3,1,4}, {3,4,1}, {3,2,4}, {3,4,2}, {4,1,2}, {4,2,1}, {4,1,3}, {4,3,1}, {4,2,3}, {4,3,2}

Permutation Problem 2

Choose 3 contestants from group of 12 contestants

At a high school track meet the 400 meter race has 12 contestants. The top 3 will receive points for their team. How many different permutations are there for the top 3 from the 12 contestants?

For this problem we are looking for an ordered subset 3 contestants (r) from the 12 contestants (n). We must calculate P(12,3) in order to find the total number of possible outcomes for the top 3.

P(12,3) = 12! / (12-3)! = 1,320 Possible Outcomes

Permutation Problem 3

Choose 5 players from a set of 10 players

An NFL team has the 6th pick in the draft, meaning there are 5 other teams drafting before them. If the team believes that there are only 10 players that have a chance of being chosen in the top 5, how many different orders could the top 5 be chosen?

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For this problem we are finding an ordered subset of 5 players (r) from the set of 10 players (n).

P(10,5)=10!/(10-5)!= 30,240 Possible Orders

References

[1] For more information on permutations and combinations please see Wolfram MathWorld: Permutation.

We can use permutations and combinations to help us answer more complex probability questions

Example 1

A 4 digit PIN is selected. What is the probability that there are no repeated digits?

There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 × 10 × 10 × 10 = 10
⁴ = 10000 total possible PINs.

To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 × 9 × 8 × 7, or notice that this is the same as the permutation
₁₀P₄ = 5040.

The probability of no repeated digits is the number of 4 digit PINs with no repeated digits divided by the total number of 4 digit PINs. This probability is

[latex]displaystylefrac{{{}_{{10}}{P}_{{4}}}}{{{10}^{{4}}}}=frac{{5040}}{{10000}}={0.504}[/latex]

Example 2

In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.

In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player’s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is
₄₈C₆ = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player’s ticket, so the probability of winning the grand prize is:

[latex]displaystylefrac{{{}_{{6}}{C}_{{6}}}}{{{}_{{48}}{C}_{{6}}}}=frac{{1}}{{12271512}}approx={0.0000000815}[/latex]

Example 3

In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.

As above, the number of possible outcomes of the lottery drawing is
₄₈C₆ = 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 ₆C₅ = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by ₄₂C₁ = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: ₆C₅ × ₄₂C₁ = 6 × 42 = 252. So the probability of winning the second prize is

[latex]displaystylefrac{{{left({}_{{6}}{C}_{{5}}right)}{left({}_{{42}}{C}_{{1}}right)}}}{{{}_{{48}}{C}_{{6}}}}=frac{{252}}{{12271512}}approx{0.0000205}[/latex]

Try it Now 1

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 exactly 9 questions correct.

Example 4

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,
₅₂C₅. 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
₄C₁ ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be ₄₈C₄ ways to select the four non-Aces. Now we use the Basic Counting Rule to calculate that there will be ₄C₁ × ₄₈C₄ ways to choose one ace and four non-Aces.

Putting this all together, we have

[latex]displaystyle{P}{left(text{one Ace}right)}=frac{{{left({}_{{4}}{C}_{{1}}right)}{left({}_{{48}}{C}_{{4}}right)}}}{{{}_{{52}}{C}_{{5}}}}=frac{{778320}}{{2598960}}approx{0.299}[/latex]

Example 5

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:

It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.

Try it Now 2

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 6

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 the second person does not share your birthday? There are 365 days in the year (let’s ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364/365. Now we move to the third person. What is the probability that this third person does not have the same birthday as either you or the second person? There are 363 days that will not duplicate your birthday or the second person’s, so the probability that the third person does not share a birthday with the first two is 363/365.

We want the second person not to share a birthday with you
and the third person not to share a birthday with the first two people, so we use the multiplication rule:

[latex]displaystyle{P}{left(text{no shared birthday}right)}=frac{{365}}{{365}}cdotfrac{{364}}{{365}}cdotfrac{{363}}{{365}}approx{0.9918}[/latex]

and then subtract from 1 to get

P(shared birthday) = 1 – P(no shared birthday) = 1 – 0.9918 = 0.0082.

This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small. Let’s make our group a bit bigger.

Example 7

Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people?

Continuing the pattern of the previous example, the answer should be

[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{365}}{{365}}cdotfrac{{364}}{{365}}cdotfrac{{363}}{{365}}cdotfrac{{362}}{{365}}cdotfrac{{361}}{{365}}approx{0.0271}[/latex]

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Note that we could rewrite this more compactly as

[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{{}_{{365}}{P}_{{5}}}}{{365}^{{5}}}approx{0.0271}[/latex]

which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.

Example 8

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

[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{{}_{{365}}{P}_{{30}}}}{{365}^{{30}}}approx{0.706}[/latex]

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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!

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 3

Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?

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1. [latex]displaystyle{P}{left({9} text{ answers correct}right)}=frac{9cdot4}{(5^{10})}approx0.0000037[/latex] chance
2. [latex]displaystyle{P}{left(text{three Aces and two Kings}right)}=frac{{{left({}_{{4}}{C}_{{3}}right)}{left({}_{{4}}{C}_{{2}}right)}}}{{{}_{{52}}{C}_{{5}}}}=frac{{24}}{{2598960}}approx{0.0000092}[/latex]
3. [latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{{}_{{365}}{P}_{{10}}}}{{365}^{{10}}}approx{0.117}[/latex]

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David Lippman, Math in Society, “Probability,” licensed under a CC BY-SA 3.0 license.