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# Permutations and Combinations Compared

## Order is important when picking r at a time from a group of n:  nPr = n!/ (n-r)!

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Permutations

You have six photographs and you want to choose three to position in three different picture frames. You're planning to hang the picture frames horizontally on your wall. How many different possible arrangements could you create? Do you think there would be at least 25 different ways? Would you be surprised to find out that there are over 100 different ways?

### Guidance

Permutations and combinations are the next step in the learning of probability. It is by using permutations and combinations that we can find the probabilities of various events occurring at the same time, such as choosing 3 boys and 3 girls from a class of grade 12 math students.

In mathematics, we use more precise language:

If the order doesn't matter, it is a combination.

If the order does matter, it is a permutation.

Say, for example, you are making a salad. You throw in some lettuce, carrots, cucumbers, and green peppers. The order in which you throw in these vegetables doesn't really matter. Here we are talking about a combination. For combinations, you are merely selecting. Say, though, that Jack went to the ATM to get out some money and that he has to put in his PIN number. Here the order of the digits in the PIN number is quite important. In this case, we are talking about a permutation. Maybe Jack goes to PE class and needs to get into his locker. The lock on the front requires entering a series of numbers in a very specific order; therefor it is a permutation lock. For permutations, you are ordering objects in a specific manner.

The Fundamental Counting Principle states that if an event can be chosen in p\begin{align*}p\end{align*} different ways and another independent event can be chosen in q\begin{align*}q\end{align*} different ways, the number of different ways the 2 events can occur is p×q\begin{align*}p \times q\end{align*}. In other words, the Fundamental Counting Principle tells you how many ways you can arrange items. Permutations are the number of possible arrangements in an ordered set of objects.

#### Example A

How many ways can you arrange the letters in the word MATH?

You have 4 letters in the word, and you are going to choose 1 letter at a time. When you choose the first letter, you have 4 possibilities ('M', 'A', 'T', or 'H'). Your second choice will have 3 possibilities, your third choice will have 2 possibilities, and your last choice will have only 1 possibility.

Therefore, the number of arrangements is: 4×3×2×1=24\begin{align*}4 \times 3 \times 2 \times 1 = 24\end{align*} possible arrangements.

The notation for a permutation is: nPr\begin{align*}{_n}P_r\end{align*},

where:

n\begin{align*}n\end{align*} is the total number of objects.

r\begin{align*}r\end{align*} is the number of objects chosen.

For simplifying calculations, when n=r\begin{align*}n = r\end{align*}, then nPr=n!\begin{align*}{_n}P_r = n!\end{align*}.

The factorial function (!) requires us to multiply a series of descending natural numbers.

Examples:

5!4!1!=5×4×3×2×1=120=4×3×2×1=24=1\begin{align*}5! &= 5 \times 4 \times 3 \times 2 \times 1 = 120\\ 4! &= 4 \times 3 \times 2 \times 1 = 24\\ 1! &= 1\end{align*}

Note: It is a general rule that 0!=1\begin{align*}0! = 1\end{align*}.

#### Example B

Solve for 4P4\begin{align*}{_4}P_4\end{align*}.

4P4=4321=24\begin{align*}{_4}P_4 = 4 \cdot 3 \cdot 2 \cdot 1 = 24\end{align*}

This represents the number of ways to arrange 4 objects that are chosen from a set of 4 different objects.

#### Example C

Solve for 6P3\begin{align*}{_6}P_3\end{align*}.

Starting with 6, multiply the first 3 numbers of the factorial:

6P3=654=120\begin{align*}{_6}P_3 = 6 \cdot 5 \cdot 4 = 120\end{align*}

This represents the number of ways to arrange 3 objects that are chosen from a set of 6 different objects.

The formula to solve permutations like these is:

nPr=n!(nr)!\begin{align*}{_n}P_r=\frac{n!}{(n-r)!}\end{align*}

Look at Example C above. In this example, the total number of objects (n)\begin{align*}(n)\end{align*} is 6, while the number of objects chosen (r)\begin{align*}(r)\end{align*} is 3. We can use these 2 numbers to calculate the number of possible permutations (or the number of arrangements) of 6 objects chosen 3 at a time.

nPr6P36P36P36P3=n!(nr)!=6!(63)!=6!3!=6×5×4×3×2×13×2×1=1201=120\begin{align*}{_n}P_r &= \frac{n!}{(n-r)!}\\ {_6}P_3 &= \frac{6!}{(6-3)!}\\ {_6}P_3 &= \frac{6!}{3!}=\frac{6 \times 5 \times 4 \cancel{\times 3 \times 2 \times 1}}{\cancel{3 \times 2 \times 1}}\\ {_6}P_3 &= \frac{120}{1}\\ {_6}P_3 &= 120\end{align*}

Now that you have seen the Permutation formula in action, from here forward it is acceptable, and highly suggested, that you use your graphing calculator as a shortcut.

### Vocabulary

The Fundamental Counting Principle states that if an event can be chosen in p\begin{align*}p\end{align*} different ways and another independent event can be chosen in q\begin{align*}q\end{align*} different ways, the probability of the 2 events occurring is p×q\begin{align*}p \times q\end{align*}. The number of possible arrangements (nPr)\begin{align*}({_n}P_r)\end{align*} in an ordered set of objects, where n=\begin{align*}n =\end{align*} the number of objects and r=\begin{align*}r =\end{align*} the number of objects selected, is the number of permutations

### Guided Practice

a. What is the total number of possible 4-letter arrangements of the letters 's', 'n', 'o', and 'w' if each letter is used only once in each arrangement?

b. A committee is to be formed with a president, a vice president, and a treasurer. If there are 10 people to select from, how many committees are possible?

a. In this problem, there are 4 letters to choose from, so n=4\begin{align*}n = 4\end{align*}. We want 4-letter arrangements; therefore, we are choosing 4 objects at a time. In this example, r=4\begin{align*}r = 4\end{align*}.

b. In this problem, there are 10 committee members to choose from, so n=10\begin{align*}n = 10\end{align*}. We want to choose 3 members to be president, vice-president, and treasurer; therefore, we are choosing 3 objects at a time. In this example, r=3\begin{align*}r = 3\end{align*}.

nPr10P310P310P310P3=n!(nr)!=10!(103)!=10!7!=10×9×8×7×6×5×4×3×2×17×6×5×4×3×2×1=7201=720\begin{align*}{_n}P_r &= \frac{n!}{(n-r)!}\\ {_{10}}P_3 &= \frac{10!}{(10-3)!}\\ {_{10}}P_3 &= \frac{10!}{7!}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\\ {_{10}}P_3 &= \frac{720}{1}\\ {_{10}}P_3 &= 720\end{align*}

### Practice

1. Solve for 7P5\begin{align*}{_7}P_5\end{align*}.
2. Evaluate 4P2×5P3\begin{align*}{_4}P_2 \times {_5}P_3\end{align*}.
3. How many different 4-digit numerals can be made from the digits of 56987 if a digit can appear just once in a numeral?
4. In how many ways can 6 students be chosen from 9 students if the order in which the students are chosen matters?
5. A TV station has 8 hour-long TV shows to choose from in order to fill 2 one-hour time slots. In how many ways can it fill the time slots?
6. A basketball league consists of 12 teams. In how many ways can the teams finish in first, second, and third place?
7. A secret code consist of 3 digits from 0 to 9 followed by 2 letters from 'A' to 'Z'. None of the digits or letters repeat. How many secret codes are possible?
8. On a test, Robert has been presented with 15 vocabulary words and 15 definitions. He is being asked to match each vocabulary word to the appropriate definition. How many different ways are there for Robert to do the matching?
9. A couple just had twin boys, but they can't decide between the names Mike, Mark, Peter, Paul, Sam, and Sonny. If the couple randomly chooses names for the 2 boys from the names listed, what is the probability that the first boy born will be named Sam and the second boy born will be named Mark? Assume that the boys will not have the same name.
10. For question 9, what is the probability that one of the boys will be named Peter and the other boy will be named Paul?

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