7.3 Permutations

Learning Objectives

In this section, you will learn to:
  1. Count the number of permutations (ordered arrangements) of \(n\) items taken \(r\) at a time.
  2. Count permutations when extra conditions are imposed on the arrangements.
  3. Perform calculations using factorials.

7.3.1 What Is a Permutation?

In Example 7.2.6 we counted the ordered arrangements of the letters \(\{A, B, C\}\) when no letter was repeated. The tree diagram gave us six arrangements:

$$\text{ABC, ACB, BAC, BCA, CAB, CBA.}$$

Arrangements like these — where order matters and no element repeats — are called permutations.

Definition 7.10: Permutation

Source: Applied Finite Mathematics

A permutation of a set of elements is an ordered arrangement in which each element is used exactly once.

Context Pause: "Order matters" is the whole game

Counting techniques split into two big families: ones where order matters (permutations) and ones where it doesn't (combinations, coming in section 7.5). "ABC" and "CBA" are the same combination of letters but different permutations. Whenever a problem asks for "arrangements," "orderings," "sequences," or anything where swapping two items creates a new answer, you are counting permutations.

Insight Note: Permutations are the multiplication axiom with a shrinking pool

Every permutation problem you have seen so far has used the same pattern: \(n\) choices for the first slot, \(n-1\) for the second (because one is now used), \(n-2\) for the third, and so on. The multiplication axiom is the permutation formula — nothing new, just a name for a common case.

Example 7.3.1

Source: Applied Finite Mathematics

How many three-letter word sequences can be formed using the letters \(\{A, B, C, D\}\)?

Solution

Step 1 — Track the shrinking pool. Four choices for the first letter, three for the second (one is now used), two for the third.

Slot 1st 2nd 3rd
Choices 4 3 2

Step 2 — Multiply.

$$4 \cdot 3 \cdot 2 = 24 \text{ arrangements.}$$

Answer: \(24\) three-letter sequences.

Try It Now 7.3.1

Source: Applied Finite Mathematics

How many four-letter arrangements can be formed from the letters \(\{W, X, Y, Z\}\) if no letter repeats?

Solution

All four letters placed in four slots, no repetition:

$$4 \cdot 3 \cdot 2 \cdot 1 = 24 \text{ arrangements.}$$

7.3.2 Permutations with Restrictions

Many real problems add a twist: a letter must be in a certain slot, two people must sit together, the first item must be a vowel. The trick is to place the restricted slots first, then fill the remaining slots with the multiplication axiom.

Handle restrictions before general choices

If a slot has a rule attached ("must be a consonant," "must be odd"), fill that slot first. Once the pool for that slot is resolved, the other slots draw from the remaining items as usual. Trying to fill restricted slots last is how double-counting errors sneak in.

Restrictions shrink the front, not the back

The restriction usually lives on a specific position (first, last, middle). After you handle that position, the remaining arrangement is still a straight shrinking-pool permutation. So a restricted problem is always a small permutation problem glued to a larger one by the multiplication axiom.

Example 7.3.2

Source: Applied Finite Mathematics

How many permutations of the letters of the word ARTICLE have consonants in the first and last positions?

Solution

Step 1 — Identify the restricted pool. ARTICLE has \(7\) letters, \(4\) of which are consonants: R, T, C, L.

Step 2 — Fill the restricted slots first. The first slot must be one of the \(4\) consonants. Once one is chosen, \(3\) consonants remain for the last slot.

Step 3 — Fill the middle slots. After placing two letters, \(5\) letters remain for the \(5\) middle positions. Multiplication axiom: \(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\).

Slot 1st 2nd 3rd 4th 5th 6th 7th
Choices 4 5 4 3 2 1 3

Step 4 — Multiply.

$$4 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 3 = 1{,}440.$$

Answer: \(1{,}440\) permutations.

Example 7.3.3

Source: Applied Finite Mathematics

Given five letters \(\{A, B, C, D, E\}\), find:

a. The number of four-letter word sequences.

b. The number of three-letter word sequences.

c. The number of two-letter word sequences.

Solution

Each part is a straight shrinking-pool permutation starting from a pool of \(5\) letters.

(a) Four-letter sequences.

$$5 \cdot 4 \cdot 3 \cdot 2 = 120.$$

(b) Three-letter sequences.

$$5 \cdot 4 \cdot 3 = 60.$$

(c) Two-letter sequences.

$$5 \cdot 4 = 20.$$

Answer: (a) \(120\); (b) \(60\); (c) \(20\).

Try It Now 7.3.2

Source: Applied Finite Mathematics

How many permutations of the letters in PLANET have a vowel in the first position?

Solution

Step 1 — Count the restricted pool. PLANET has \(2\) vowels (A, E).

Step 2 — Place the first slot. \(2\) choices.

Step 3 — Fill the remaining \(5\) slots. \(5\) letters remain: \(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\).

Step 4 — Multiply.

$$2 \cdot 120 = 240 \text{ permutations.}$$

7.3.3 Factorials and the \(nPr\) Formula

When we count permutations of \(n\) objects taken \(r\) at a time, it's convenient to have a single symbol for the answer. That symbol is \(nPr\) (sometimes written \({}_nP_r\) or \(P(n, r)\)).

Definition 7.11: Factorial

Definition 7.11: Factorial

Source: Applied Finite Mathematics

For a natural number \(n\), the factorial of \(n\), written \(n!\), is the product of all positive integers from \(n\) down to \(1\):

$$n! = n (n - 1)(n - 2)(n - 3) \cdots 3 \cdot 2 \cdot 1.$$

By convention, \(0! = 1\).

Definition 7.12: Permutations of \(n\) Objects Taken \(r\) at a Time

Definition 7.12: Permutations of \(n\) Objects Taken \(r\) at a Time

Source: Applied Finite Mathematics

The number of permutations of \(n\) objects taken \(r\) at a time, written \(nPr\), is

$$nPr = n(n - 1)(n - 2) \cdots (n - r + 1),$$

or equivalently,

$$nPr = \dfrac{n!}{(n - r)!}.$$

Both formulas give the same answer; become comfortable with both.

Context Pause: Why two formulas for the same thing?

The product form \(n(n-1)\cdots(n-r+1)\) is usually easier to compute by hand — you stop multiplying after \(r\) terms. The factorial-quotient form \(\dfrac{n!}{(n-r)!}\) matches how calculators and statistical software compute it internally, and it makes the connection to combinations (section 7.5) clean. Different tools, same idea.

Insight Note: \(0! = 1\) is a convenience, not a mystery

Every formula that uses factorials — \(nPr\), \(nCr\), the binomial theorem — needs \(0! = 1\) so that edge cases work out correctly. It isn't a mathematical accident; it's the definition that keeps the algebra clean. Accept it and move on.

Example 7.3.4

Example 7.3.4

Source: Applied Finite Mathematics

Compute the following using both formulas.

a. \(6P3\)

b. \(7P2\)

Solution

(a) \(6P3\).

Product form: \(n = 6\), \(r = 3\), so multiply three terms starting at \(6\):

$$6P3 = 6 \cdot 5 \cdot 4 = 120.$$

Factorial form:

$$6P3 = \dfrac{6!}{(6-3)!} = \dfrac{6!}{3!} = \dfrac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1} = 120.$$

(b) \(7P2\).

Product form: \(n = 7\), \(r = 2\):

$$7P2 = 7 \cdot 6 = 42.$$

Factorial form:

$$7P2 = \dfrac{7!}{5!} = \dfrac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 42.$$

Answer: (a) \(120\); (b) \(42\).

Try It Now 7.3.3

Try It Now 7.3.3

Source: Applied Finite Mathematics

Compute \(8P4\) using the product form.

Solution

\(n = 8\), \(r = 4\), multiply four terms starting at \(8\):

$$8P4 = 8 \cdot 7 \cdot 6 \cdot 5 = 1{,}680.$$

7.3.4 Combined Arrangements

Sometimes an arrangement has two independent pieces — and each piece is itself a permutation. The multiplication axiom lets us combine them: count each piece on its own, then multiply.

Combine permutations by multiplying

If a problem splits naturally into independent parts ("math books in slots 1–3, history books in slots 4–5"), handle each part as its own permutation and multiply the results. This is the same multiplication axiom we have been using all along, just applied to permutation counts instead of raw choice counts.

"Glue two items together" trick

When two specific items must sit next to each other, treat them as a single combined item. That reduces the problem from \(n\) items to \(n-1\) items. Then multiply by the number of ways the two glued items can be ordered relative to each other (usually \(2!\)).

Example 7.3.5

Source: Applied Finite Mathematics

In how many different ways can \(4\) people be seated in a straight line if two of them insist on sitting next to each other?

Solution

Step 1 — Glue the two together. Call the people A, B, C, D, and suppose A and B must sit together. Treat AB as a single combined "super-person," leaving three items: AB, C, D.

Step 2 — Arrange the three items. Three items in three seats: \(3! = 6\) orderings.

Step 3 — Account for the internal order of AB. A and B can sit either as AB or BA — \(2!\) internal orderings.

Step 4 — Multiply.

$$3! \cdot 2! = 6 \cdot 2 = 12.$$

Sanity check — enumerate. With AB glued: ABCD, ABDC, CABD, DABC, CDAB, DCAB. With BA glued: BACD, BADC, CBAD, DBAC, CDBA, DCBA. That's \(12\) arrangements — match.

Answer: \(12\) seatings.

Example 7.3.6

Source: Applied Finite Mathematics

You have \(4\) math books and \(5\) history books to put on a shelf that has \(5\) slots. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books?

Solution

Step 1 — Count the math-book arrangements. Three slots from \(4\) math books: \(4P3 = 4 \cdot 3 \cdot 2 = 24\).

Step 2 — Count the history-book arrangements. Two slots from \(5\) history books: \(5P2 = 5 \cdot 4 = 20\).

Step 3 — Multiply. Every math-book arrangement pairs with every history-book arrangement.

$$(4P3)(5P2) = 24 \cdot 20 = 480.$$

Slot1st2nd3rd4th5th
Choices43254

Answer: \(480\) arrangements.

Try It Now 7.3.4

Source: Applied Finite Mathematics

In how many ways can \(6\) people be seated in a row if three specific people must sit together?

Solution

Step 1 — Glue the three together. Treat the trio as a single super-person, leaving \(4\) items to arrange.

Step 2 — Arrange the four items. \(4! = 24\).

Step 3 — Account for the internal order of the trio. \(3! = 6\) orderings within the glued group.

Step 4 — Multiply.

$$4! \cdot 3! = 24 \cdot 6 = 144 \text{ seatings.}$$

Problem Set 7.3

Source: Applied Finite Mathematics

Problems 1–8: Do the following problems using permutations.

Problem 1. How many three-letter words can be made using the letters \(\{a, b, c, d, e\}\) if no repetitions are allowed?

Problem 1 Solution

Step 1 — Shrinking pool. Three slots from \(5\) letters, no repetition:

$$5 \cdot 4 \cdot 3 = 60.$$

Answer: \(60\) words.

Problem 2. A grocery store has five checkout counters and seven clerks. How many different ways can the 7 clerks be assigned to the 5 counters?

Problem 2 Solution

Step 1 — Order matters. Assign \(5\) of the \(7\) clerks to \(5\) distinct counters:

$$7P5 = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 = 2{,}520.$$

Answer: \(2{,}520\) assignments.

Problem 3. A group of fifteen people who are members of an investment club wish to choose a president and a secretary. How many different ways can this be done?

Problem 3 Solution

Step 1 — Two ordered positions from \(15\).

$$15P2 = 15 \cdot 14 = 210.$$

Answer: \(210\) ways.

Problem 4. Compute the following.

  1. \(9P2\)
  2. \(6P4\)
  3. \(8P3\)
  4. \(7P4\)
Problem 4 Solution

(a) \(9P2\): \(9 \cdot 8 = 72\).

(b) \(6P4\): \(6 \cdot 5 \cdot 4 \cdot 3 = 360\).

(c) \(8P3\): \(8 \cdot 7 \cdot 6 = 336\).

(d) \(7P4\): \(7 \cdot 6 \cdot 5 \cdot 4 = 840\).

Problem 5. In how many ways can the letters of the word CUPERTINO be arranged if each letter is used only once in each arrangement?

Problem 5 Solution

Step 1 — Count distinct letters. CUPERTINO has \(9\) distinct letters: C, U, P, E, R, T, I, N, O.

Step 2 — Arrange all \(9\) in \(9\) slots.

$$9! = 362{,}880.$$

Answer: \(362{,}880\) arrangements.

Problem 6. How many permutations of the letters of the word PROBLEM end in a vowel?

Problem 6 Solution

Step 1 — Identify the restricted pool. PROBLEM has \(7\) distinct letters; vowels are O and E (\(2\) vowels).

Step 2 — Fill the last slot. \(2\) choices.

Step 3 — Fill the remaining \(6\) slots. \(6! = 720\).

Step 4 — Multiply.

$$2 \cdot 720 = 1{,}440.$$

Answer: \(1{,}440\) permutations.

Problem 7. How many permutations of the letters of the word SECURITY end in a consonant?

Problem 7 Solution

Step 1 — Identify the restricted pool. SECURITY has \(8\) distinct letters; consonants are S, C, R, T, Y (\(5\) consonants).

Step 2 — Fill the last slot. \(5\) choices.

Step 3 — Fill the remaining \(7\) slots. \(7! = 5{,}040\).

Step 4 — Multiply.

$$5 \cdot 5{,}040 = 25{,}200.$$

Answer: \(25{,}200\) permutations.

Problem 8. How many permutations of the letters PRODUCT have consonants in the second and third positions?

Problem 8 Solution

Step 1 — Identify the restricted pool. PRODUCT has \(7\) distinct letters; consonants are P, R, D, C, T (\(5\) consonants).

Step 2 — Fill slots 2 and 3 with consonants. \(5\) choices for slot 2, \(4\) remaining for slot 3.

Step 3 — Fill the other \(5\) slots. \(5! = 120\).

Step 4 — Multiply.

$$5 \cdot 4 \cdot 120 = 2{,}400.$$

Answer: \(2{,}400\) permutations.

Problems 9–16: Do the following problems using permutations.

Problem 9. How many three-digit numbers are there?

Problem 9 Solution

Step 1 — First digit can't be \(0\). \(9\) choices for the hundreds place, \(10\) for tens, \(10\) for units:

$$9 \cdot 10 \cdot 10 = 900.$$

Answer: \(900\) three-digit numbers.

Problem 10. How many three-digit odd numbers are there?

Problem 10 Solution

Step 1 — Last digit must be odd. \(5\) choices: \(\{1, 3, 5, 7, 9\}\).

Step 2 — First digit can't be \(0\). \(9\) choices.

Step 3 — Tens digit has no restriction. \(10\) choices.

Step 4 — Multiply.

$$9 \cdot 10 \cdot 5 = 450.$$

Answer: \(450\) three-digit odd numbers.

Problem 11. In how many different ways can five people be seated in a row if two of them insist on sitting next to each other?

Problem 11 Solution

Step 1 — Glue the two together. Five people with two tied becomes \(4\) items: \(4! = 24\).

Step 2 — Internal order of the tied pair. \(2! = 2\).

Step 3 — Multiply.

$$4! \cdot 2! = 24 \cdot 2 = 48.$$

Answer: \(48\) seatings.

Problem 12. In how many different ways can five people be seated in a row if two of them insist on not sitting next to each other?

Problem 12 Solution

Step 1 — Total arrangements. \(5! = 120\).

Step 2 — Subtract the together-arrangements from Problem 11. \(120 - 48 = 72\).

Answer: \(72\) seatings where the two do not sit together.

Problem 13. In how many ways can 3 English, 3 history, and 2 math books be set on a shelf if the English books are set on the left, history books in the middle, and math books on the right?

Problem 13 Solution

Step 1 — Order inside each group. Subjects are fixed in place (E left, H middle, M right), so only internal orderings matter.

  • English: \(3! = 6\)
  • History: \(3! = 6\)
  • Math: \(2! = 2\)

Step 2 — Multiply.

$$6 \cdot 6 \cdot 2 = 72.$$

Answer: \(72\) arrangements.

Problem 14. In how many ways can 3 English, 3 history, and 2 math books be set on a shelf if they are grouped by subject?

Problem 14 Solution

Step 1 — Order the three subject groups. \(3! = 6\).

Step 2 — Internal orderings. \(3! \cdot 3! \cdot 2! = 72\) (from Problem 13).

Step 3 — Multiply.

$$3! \cdot 72 = 6 \cdot 72 = 432.$$

Answer: \(432\) arrangements.

Problem 15. You have 5 math books and 6 history books to put on a shelf with five slots. In how many ways can you put the books on the shelf if the first two slots are to be filled with math books and the next three with history books?

Problem 15 Solution

Step 1 — First two slots from \(5\) math books. \(5P2 = 5 \cdot 4 = 20\).

Step 2 — Next three slots from \(6\) history books. \(6P3 = 6 \cdot 5 \cdot 4 = 120\).

Step 3 — Multiply.

$$20 \cdot 120 = 2{,}400.$$

Answer: \(2{,}400\) arrangements.

Problem 16. You have 5 math books and 6 history books to put on a shelf with five slots. In how many ways can you put the books on the shelf if the first two slots are to be filled with the books of one subject and the next three slots are to be filled with the books of the other subject?

Problem 16 Solution

Case A — First 2 math, next 3 history. \(5P2 \cdot 6P3 = 20 \cdot 120 = 2{,}400\) (same as Problem 15).

Case B — First 2 history, next 3 math. \(6P2 \cdot 5P3 = 30 \cdot 60 = 1{,}800\).

Add the cases.

$$2{,}400 + 1{,}800 = 4{,}200.$$

Answer: \(4{,}200\) arrangements.

Problems 17–22: Do the following problems using permutations.

Problem 17. A bakery has 9 different fancy cakes. In how many ways can 5 of the 9 fancy cakes be lined up in a row in the bakery display case?

Problem 17 Solution

Step 1 — Five of nine ordered. \(9P5 = 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 = 15{,}120\).

Answer: \(15{,}120\) displays.

Problem 18. A landscaper has 6 different flowering plants. She needs to plant 4 of them in a row in a garden. How many different ways can 4 of the 6 plants be arranged in a row?

Problem 18 Solution

Step 1 — Four of six ordered. \(6P4 = 6 \cdot 5 \cdot 4 \cdot 3 = 360\).

Answer: \(360\) arrangements.

Problem 19. At an auction of used construction vehicles, there are 7 different vehicles for sale. In how many orders could these 7 vehicles be listed in the auction program?

Problem 19 Solution

Step 1 — All seven ordered. \(7! = 5{,}040\).

Answer: \(5{,}040\) orders.

Problem 20. A landscaper has 6 different flowering plants and 4 different non-flowering bushes. She needs to plant a row of 6 plants in a garden. There must be a bush at each end, and four flowering plants in a row in between the bushes. How many different arrangements in a row are possible?

Problem 20 Solution

Step 1 — Bushes at the two ends. \(4P2 = 4 \cdot 3 = 12\).

Step 2 — Four flowers in the middle four slots. \(6P4 = 6 \cdot 5 \cdot 4 \cdot 3 = 360\).

Step 3 — Multiply.

$$12 \cdot 360 = 4{,}320.$$

Answer: \(4{,}320\) arrangements.

Problem 21. In how many ways can all 7 letters of the word QUIETLY be arranged if the letters Q and U must be next to each other in the order QU?

Problem 21 Solution

Step 1 — Glue QU as one item. QUIETLY has \(7\) letters; gluing Q and U gives \(6\) items to arrange.

Step 2 — Arrange six items. \(6! = 720\).

Step 3 — Fix internal order. Q must come before U, so no \(2!\) multiplier.

Answer: \(720\) arrangements.

Problem 22. In how many ways can the letters ABCDEXY be arranged if

a) the X and Y must be next to each other in either order (XY or YX)?

b) the X and Y cannot be next to each other?

Problem 22 Solution

(a) X and Y together, either order. Glue XY as one item: \(6\) items, \(6! = 720\). Two internal orderings (XY or YX): \(2! = 2\). Total:

$$6! \cdot 2! = 720 \cdot 2 = 1{,}440.$$

(b) X and Y not together. All permutations of \(7\) distinct letters \(= 7! = 5{,}040\). Subtract the together-arrangements from part (a):

$$5{,}040 - 1{,}440 = 3{,}600.$$

Answer: (a) \(1{,}440\); (b) \(3{,}600\).