Add Maths · 11 Permutations and combinations

11 Permutations and combinations

Before you start

Multiplying whole numbers and cancelling common factors in a fraction without a calculator.
Expanding a product such as $n (n - 1) (n - 2)$ and solving a quadratic equation by factorising.
Listing outcomes in a systematic order, for example all the orders of the letters A, B and C.

Syllabus objectives covered (3)

Recognise the difference between permutations and combinations and know when each should be used.
Know and use the notation n! and the expressions for permutations and combinations of n items taken r at a time.
Solve problems on arrangement and selection using permutations or combinations.

Part 1 of 8

Counting by filling positions

Every count in this topic is built from one rule. If one choice can be made in $a$ ways and, after it, a second choice can be made in $b$ ways, then the two choices together can be made in $a \times b$ ways.
Multiply, do not add, because each of the $a$ first choices can be followed by every one of the $b$ second choices. Three shirts and four pairs of trousers make $3 \times 4 = 12$ outfits, not $7$ .
Now arrange 4 different books in a row on a shelf. Fill the positions from left to right. Any of the 4 books can go first. Once one is placed, 3 are left for the second position, then 2 for the third, then 1 for the last.
$4 \times 3 \times 2 \times 1 = 24$
The numbers go down by one each time because a book that is already on the shelf cannot be used again.
The syllabus leaves out problems with repeated objects (such as the two Ts in LETTER) and objects arranged in a circle. So every object you count in this topic is different from the others, and each one is used at most once.

Part 2 of 8

Factorial notation, and why 0! = 1

The product $4 \times 3 \times 2 \times 1$ has its own symbol, $4!$ , read as '4 factorial'. For any positive whole number $n$ :
$n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1$
So $n!$ is the number of ways to arrange $n$ different objects in a row. Know the first few on sight: $1! = 1$ , $2! = 2$ , $3! = 6$ , $4! = 24$ , $5! = 120$ , $6! = 720$ , $7! = 5040$ .
The fact that makes factorials easy to handle: each factorial contains the one below it.
$n! = n \times (n - 1)!$
Use it to divide factorials without a calculator. Write out the larger factorial only until you reach the smaller one, then cancel.
$\frac{10!}{7!} = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8 = 720$
Do not cancel the numbers in front of the $!$ as if they were ordinary numbers. $\frac{10!}{5!}$ is not $2!$ . It is $10 \times 9 \times 8 \times 7 \times 6 = 30 240$ .
$0!$ is defined to be $1$ . Put $n = 1$ into $n! = n \times (n - 1)!$ : it gives $1! = 1 \times 0!$ , and $1! = 1$ , so $0!$ must be $1$ . This is the value that makes the formulas below give the right answer when you choose none of the objects or all of them.
The binomial theorem on the List of formulas includes $(\binom{n}{r}) = \frac{n!}{(n - r)! r!}$ , but there is no formula for $^{n} P_{r}$ , and topic 11 has no note saying its formulas are given. So learn $n!$ and both expressions in the next two sections. Paper 1 does not allow a calculator, so practise evaluating them by cancelling.

Part 3 of 8

Permutations: when order matters

A permutation is an arrangement in which the order counts. 'Ann first, Ben second' and 'Ben first, Ann second' are two different permutations.
Eight runners finish a race. In how many ways can first, second and third place be filled? Fill the three positions in turn: 8 choices for first, then 7 for second, then 6 for third.
$8 \times 7 \times 6 = 336$
The notation $^{n} P_{r}$ means the number of permutations of $r$ objects taken from $n$ different objects. You have just found $^{8} P_{3} = 336$ . In general it is $r$ factors, counting down from $n$ :
$^{n} P_{r} = n (n - 1) (n - 2) \dots (n - r + 1)$
To write it with factorials, notice that $8 \times 7 \times 6$ is $8!$ with the tail $5 \times 4 \times 3 \times 2 \times 1 = 5!$ removed. So $^{8} P_{3} = \frac{8!}{5!}$ , and in general:
$^{n} P_{r} = \frac{n!}{(n - r)!}$
Check the extreme case. Arranging all $n$ objects means $r = n$ , and the formula gives $\frac{n!}{0!} = \frac{n!}{1} = n!$ . That matches the $4!$ ways to arrange 4 books. If $0!$ were $0$ , the formula would break here.
If your calculator has an nPr key, you may use it on Paper 2. Still write the expression, such as $^{8} P_{3}$ , before the value: the expression is your method.

Part 4 of 8

Combinations: when order does not matter

A combination is a selection in which the order does not count. Choosing Ann, Ben and Cara for a team gives the same team whichever of them you pick first.
Count the teams of 3 that can be chosen from the 8 runners. Start from the $336$ ordered choices. Each team of 3 appears in that list once for every order of its members, and 3 people can be put in order in $3! = 6$ ways: ABC, ACB, BAC, BCA, CAB, CBA.
So every team has been counted 6 times. Divide by 6 to count each team once.
$^{8} C_{3} = \frac{^{8} P_{3}}{3!} = \frac{336}{6} = 56$
The same argument works for any $n$ and $r$ : divide the number of permutations by $r!$ , the number of orders of the $r$ chosen objects.
$^{n} C_{r} = (\binom{n}{r}) = \frac{n!}{r! (n - r)!}$
$^{n} C_{r}$ and $(\binom{n}{r})$ are two ways of writing the same number. $(\binom{n}{r})$ is the form used in the binomial theorem.
By hand, cancel the larger factorial in the denominator first, so only $r$ factors are left on top.
$^{10} C_{4} = \frac{10!}{4! 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24} = 210$
Choosing 3 people to take is the same as choosing the 5 to leave behind, so $^{8} C_{3} =^{8} C_{5}$ . In general $^{n} C_{r} =^{n} C_{n - r}$ . Use whichever is quicker: $^{20} C_{18} =^{20} C_{2} = \frac{20 \times 19}{2} = 190$ .
Three values to know on sight: $^{n} C_{0} = 1$ (one way to choose nothing), $^{n} C_{n} = 1$ (one way to choose everything) and $^{n} C_{1} = n$ . The first two need $0! = 1$ in the formula.

Part 5 of 8

Permutation or combination? The swap test

Before you write any expression, ask one question: if I swap two of the chosen objects, do I get a different outcome?
Yes means order matters, so use a permutation. Examples: people in a queue, a number made from digits ( $352$ and $253$ are different numbers), first, second and third prizes, or a chair and a secretary (two different jobs).
No means order does not matter, so use a combination. Examples: a committee, a team, 3 questions chosen to answer from 8, a hand of cards, or toppings on a pizza.
Do not decide from a single word. 'Choose' appears in both kinds of question. 'Choose a chair and a secretary' is a permutation, because the two jobs are different. Look at whether the outcome has positions or roles.
Check your answer with this fact: for the same $n$ and $r$ , the number of permutations is $r!$ times the number of combinations. If an arrangement count comes out smaller than a selection count, you have used the wrong one.
The syllabus leaves out problems that need both permutations and combinations in one answer. So once you have decided which one a problem needs, count the whole problem that way.

Part 6 of 8

Arrangements with conditions

Fill the restricted positions first, then fill the others freely. If you fill the free positions first, you no longer know how many choices are left for the restricted one.
Example: four-digit numbers made from the digits 1, 2, 3, 4, 5, 6, with no digit used twice, that are even. The last digit must be 2, 4 or 6: 3 choices. Then the first three positions are filled from the 5 digits that are left.
$5 \times 4 \times 3 \times 3 = 180$
Watch for $0$ among the digits. A four-digit number cannot begin with $0$ , so the first position has one choice fewer. If $0$ can also be the last digit, the count for the first position changes with the last digit, so split into two cases (last digit $0$ , last digit not $0$ ) and add them.
Objects that must be together. Tie them into one block. Arrange the block with the other objects, then multiply by the number of orders inside the block. Six people in a row with A and B together: 5 items (the block and 4 others) give $5!$ , and the block can be AB or BA.
$5! \times 2! = 120 \times 2 = 240$
Objects that must be apart. Count the opposite and subtract it from the total. A and B not next to each other: every arrangement either has them together or does not.
$6! - 240 = 720 - 240 = 480$
Adding and multiplying have different jobs. Multiply the stages of one outcome: first digit and middle digits and last digit. Add separate cases that cannot happen at the same time: last digit $0$ or last digit not $0$ .

Part 7 of 8

Selections with conditions

Choosing from two groups. Choose from each group separately, then multiply. A committee of 2 teachers and 3 students from 6 teachers and 8 students:
$^{6} C_{2} \times^{8} C_{3} = 15 \times 56 = 840$
Multiply because any pair of teachers can sit with any set of three students.
A named person must be in. Put them in first, then choose the other $r - 1$ members from the other $n - 1$ people: $^{n - 1} C_{r - 1}$ . A named person must be out. Remove them and choose all $r$ from the other $n - 1$ : $^{n - 1} C_{r}$ .
'At least' and 'at most'. Either list the allowed cases and add them, or subtract the cases you do not want from the total. Use the way with fewer cases.
Example: a committee of 3 from 4 men and 3 women, with at least one woman. The only case you do not want is no women at all, which is 3 men chosen from 4. Subtract it from the total.
$^{7} C_{3} -^{4} C_{3} = 35 - 4 = 31$
The trap here: 'pick one woman first, then any two of the other six' gives $3 \times^{6} C_{2} = 45$ , which is too many. A committee with two women is counted twice, once for each woman who could have been 'picked first'.
Two people who must not both be chosen. Count the selections that contain both of them, then subtract that from the total.

Part 8 of 8

Algebraic problems: finding n

Some questions give the size of a count and ask for $n$ . Replace each $^{n} P_{r}$ or $^{n} C_{r}$ with its product of factors, then solve the equation.
Write the small cases as expressions in $n$ . Each one comes from the factorial form by cancelling: for example $\frac{n!}{(n - 2)!} = n (n - 1)$ , because $n! = n (n - 1) \times (n - 2)!$ .
$^{n} P_{2} = n (n - 1),^{n} C_{2} = \frac{n (n - 1)}{2},^{n} C_{3} = \frac{n (n - 1) (n - 2)}{6}$
Example: $^{n} P_{2} = 56$ gives $n (n - 1) = 56$ , so $n^{2} - n - 56 = 0$ and $(n - 8) (n + 7) = 0$ .
$n$ counts objects, so it must be a positive whole number, at least as big as $r$ . Reject $n = - 7$ and give $n = 8$ . Write down that you reject the negative root, and why.
If both sides share a factor such as $n (n - 1)$ , divide it out instead of expanding everything. This is allowed because $n (n - 1)$ is not zero for any $n$ the question can mean. The last worked example does this.

Worked examples

Example 1. A club has 12 members. (a) A captain, a vice-captain and a treasurer are to be chosen from the members, and no member may hold more than one job. Find the number of ways this can be done. (b) Instead, 3 members are to be chosen to go to a meeting. Find the number of ways this can be done.

Show the working

Apply the swap test to (a). If Ali is captain and Bea is treasurer, swapping them gives a different outcome. The jobs make order matter, so this is a permutation.
Fill the three jobs in turn: 12 choices for captain, then 11 for vice-captain, then 10 for treasurer.
$^{12} P_{3} = \frac{12!}{9!} = 12 \times 11 \times 10 = 1320$
Apply the swap test to (b). Three people going to a meeting are the same group whatever order you list them in, so this is a combination.
Each group of 3 was counted $3! = 6$ times in part (a), once for each way of giving its members the three jobs. Divide by 6.
$^{12} C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1320}{6} = 220$
Check: the permutation count is $3! = 6$ times the combination count, as it must be for the same $n$ and $r$ .

Answer: (a) $1320$ (b) $220$

Example 2. The 7 letters of the word PROBLEM are arranged in a row, each letter used once. Find the number of arrangements in which (a) there is no restriction, (b) the arrangement begins and ends with a vowel, (c) the two vowels are not next to each other.

Show the working

All 7 letters are different, and swapping two letters gives a new arrangement, so order matters. (a) There are $7! = 5040$ arrangements.
(b) The vowels are O and E. Fill the restricted positions first. The first letter is O or E: 2 choices. The last letter must be the other vowel: 1 choice.
The 5 consonants P, R, B, L, M fill the 5 middle positions in $5!$ ways.
$2 \times 1 \times 5! = 2 \times 120 = 240$
(c) Count the opposite first: O and E next to each other. Tie them into one block. That leaves 6 items to arrange: the block and the 5 consonants.
$6! \times 2! = 720 \times 2 = 1440$
The $2!$ is for the order inside the block, OE or EO. Leaving it out would count only half of these arrangements.
Every arrangement has the vowels either together or apart, so subtract from the total.
$5040 - 1440 = 3600$

Answer: (a) $5040$ (b) $240$ (c) $3600$

Example 3. A committee of 5 people is to be chosen from 6 teachers and 8 students. Find the number of different committees that (a) contain exactly 2 teachers, (b) contain at least 1 teacher. (c) One of the teachers and one of the students refuse to serve together. Find the number of committees that do not contain both of them.

Show the working

A committee is the same whatever order its members are chosen in, so every count here is a combination.
(a) Exactly 2 teachers means the other 3 members are students. Choose from each group, then multiply.
$^{6} C_{2} \times^{8} C_{3} = 15 \times 56 = 840$
(b) 'At least 1 teacher' covers 1, 2, 3, 4 or 5 teachers: five cases. Its opposite is one case, no teachers, so subtract that from the total.
$^{14} C_{5} -^{8} C_{5} = 2002 - 56 = 1946$
Work out $^{14} C_{5}$ by cancelling: in $\frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1}$ , the $12 \times 10 = 120$ on top cancels the $5! = 120$ underneath, leaving $14 \times 13 \times 11 = 2002$ . Also $^{8} C_{5} =^{8} C_{3} = 56$ .
(c) Count the committees that contain both of them, then subtract. If both are in, the other 3 members come from the remaining 12 people.
$^{12} C_{3} = \frac{12 \times 11 \times 10}{6} = 220$
Subtract from the total number of committees.
$2002 - 220 = 1782$

Answer: (a) $840$ (b) $1946$ (c) $1782$

Example 4. Given that $^{n} C_{3} = 4 \times^{n} C_{2}$ , find the value of $n$ .

Show the working

Write each side as a product in $n$ , using the factorial formula and cancelling.
$\frac{n (n - 1) (n - 2)}{6} = 4 \times \frac{n (n - 1)}{2}$
$^{n} C_{3}$ only has a meaning when $n \geq 3$ , so $n (n - 1)$ is not zero. Divide both sides by it. The right-hand side becomes $\frac{4}{2} = 2$ .
$\frac{n - 2}{6} = 2$
Multiply both sides by 6: $n - 2 = 12$ , so $n = 14$ .
Check by substituting back: $^{14} C_{3} = \frac{14 \times 13 \times 12}{6} = 364$ and $^{14} C_{2} = \frac{14 \times 13}{2} = 91$ . Since $4 \times 91 = 364$ , the answer is right.

Answer: $n = 14$

Where people go wrong

Mistake: Using $^{n} P_{r}$ for a committee or team, for example giving $^{8} P_{3} = 336$ as the number of teams of 3 from 8 people.

Instead: Use $^{n} C_{r}$ : $^{8} C_{3} = 56$ . Apply the swap test before you choose.

$^{n} P_{r}$ counts each group once for every order of its members, so it is $r!$ times too big when order does not matter.

Mistake: Taking $0! = 0$ .

Instead: $0! = 1$ .

$1! = 1 \times 0!$ forces $0! = 1$ . With $0! = 0$ , $^{n} P_{n} = \frac{n!}{0!}$ would mean dividing by zero.

Mistake: Adding the choices from two groups: $^{6} C_{2} +^{8} C_{3}$ for 2 teachers and 3 students.

Instead: Multiply: $^{6} C_{2} \times^{8} C_{3} = 15 \times 56 = 840$ .

Every pair of teachers can go with every set of three students. Add only for separate cases joined by 'or'.

Mistake: For 'at least one woman' on a committee of 3 from 4 men and 3 women, choosing one woman first and then any two others: $3 \times^{6} C_{2} = 45$ .

Instead: Subtract the unwanted case from the total: $^{7} C_{3} -^{4} C_{3} = 35 - 4 = 31$ .

A committee with two women is counted twice and one with three women three times, once for each woman who could have been chosen first.

Mistake: Leaving out the order inside a block: giving $5! = 120$ as the number of ways 6 people can stand in a row with A and B together.

Instead: $5! \times 2! = 240$ .

AB and BA are different arrangements of the row, and both have A and B together.

Check yourself

Work out $\frac{9!}{6! 3!}$ without a calculator.
Answer
$\frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$ . This is $^{9} C_{3}$ .
Write down the values of $0!$ and $1!$ . Use them to find $^{7} P_{7}$ and $^{7} C_{0}$ .
Answer
$0! = 1$ and $1! = 1$ . $^{7} P_{7} = \frac{7!}{0!} = 5040$ and $^{7} C_{0} = \frac{7!}{0! 7!} = 1$ .
State whether each count needs a permutation or a combination, then work it out. (a) Four-digit codes for a lock, made from the digits 0 to 9 with no digit repeated (a code may begin with 0). (b) Choosing 3 books from 10 to take on holiday. (c) Giving gold, silver and bronze medals to 3 of 9 skaters.
Answer
(a) Permutation, because 1234 and 4321 are different codes: $^{10} P_{4} = 5040$ . (b) Combination: $^{10} C_{3} = 120$ . (c) Permutation, because the medals are different: $^{9} P_{3} = 504$ .
Five different books are placed in a row on a shelf. Two particular books must be at the two ends. Find the number of arrangements.
Answer
Ends first: the two books go at the ends in $2! = 2$ ways. The other 3 books fill the middle in $3! = 6$ ways. Total $2 \times 6 = 12$ .
A team of 4 is chosen from 9 people, one of whom is Ana. Find the number of teams that include Ana, and the number that do not.
Answer
Include Ana: $^{8} C_{3} = 56$ . Leave her out: $^{8} C_{4} = 70$ . Check: $56 + 70 = 126 =^{9} C_{4}$ .
In a group of people, every pair shakes hands exactly once, and there are 66 handshakes. Explain why this is a combination count, and find the number of people.
Answer
A handshake between A and B is the same as one between B and A, so order does not matter. $^{n} C_{2} = 66$ gives $n (n - 1) = 132$ , so $n^{2} - n - 132 = 0$ and $(n - 12) (n + 11) = 0$ . There are $12$ people ( $n = - 11$ is rejected).
Show that $^{10} C_{7} =^{10} C_{3}$ without working out either value, and explain what this means in words.
Answer
$^{10} C_{7} = \frac{10!}{7! 3!}$ and $^{10} C_{3} = \frac{10!}{3! 7!}$ , which is the same expression. Choosing 7 objects to take is the same as choosing the 3 to leave. (Both equal $120$ .)

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Four-digit numbers are to be formed from the digits 0, 2, 3, 5, 7, 8 and 9. No digit may be used more than once in a number, and a number cannot begin with 0.

(a) Find the number of different four-digit numbers that can be formed. 2 marks No calculator

Mark scheme

M1 $6 \times 6 \times 5 \times 4$ oe, with 6 choices for the first digit because 0 is excluded, or $^{7} P_{4} -^{6} P_{3}$
A1 $720$ cao

(b) Find how many of these four-digit numbers are even. 3 marks No calculator

Mark scheme

M1 case last digit 0 counted as $6 \times 5 \times 4 = 120$ oe
M1 case last digit 2 or 8 counted as $2 \times 5 \times 5 \times 4 = 200$ oe, with 5 choices for the first digit
A1 $320$ cao. Alternative method: $3 \times^{6} P_{3} - 2 \times^{5} P_{2} = 360 - 40$ (all numbers ending in 0, 2 or 8, minus those that begin with 0) earns M1 M1 A1

Question 2. A team of 6 people is to be chosen from a group of 7 girls and 5 boys.

(a) Explain why the number of possible teams is counted using combinations and not permutations. 1 mark

Mark scheme

B1 the order in which the members are chosen does not matter: the same 6 people form the same team oe

(b) Find the number of different teams that can be chosen. 1 mark

Mark scheme

B1 $924$ cao (from $^{12} C_{6}$ )

(c) Find the number of teams that contain exactly 4 girls. 2 marks

Mark scheme

M1 $^{7} C_{4} \times^{5} C_{2}$ oe, e.g. $35 \times 10$
A1 $350$ cao

(d) Find the number of teams that contain at least 2 boys. 3 marks

Mark scheme

M1 total minus the teams with 0 boys and with 1 boy, or the sum of the cases with 2, 3, 4 and 5 boys
A1 correct values for their cases: $7$ and $105$ , or $350$ , $350$ , $105$ and $7$
A1 $812$ cao

(e) Two of the girls are sisters who must either both be in the team or both be left out. Find the number of teams that meet this condition. 3 marks

Mark scheme

M1 both sisters in: $^{10} C_{4} = 210$
M1 both sisters out: $^{10} C_{6} = 210$
A1 $420$ cao, from the two cases added. Alternative method: total minus teams with exactly one sister, $924 - 2 \times^{10} C_{5} = 924 - 504$ , earns M1 M1 A1

Question 3.

(a) Show that $^{n} C_{3} = \frac{n (n - 1) (n - 2)}{6}$ for $n \geq 3$ . 2 marks No calculator

Mark scheme

M1 $^{n} C_{3} = \frac{n!}{3! (n - 3)!}$
A1 $\frac{n!}{(n - 3)!} = n (n - 1) (n - 2)$ and $3! = 6$ shown, reaching the given result with no errors; nfww

(b) Hence, given that $^{n} C_{3} = 7 n$ , find the value of $n$ . 3 marks No calculator

Mark scheme

M1 $\frac{n (n - 1) (n - 2)}{6} = 7 n$ , divided by $n$ to give $(n - 1) (n - 2) = 42$ oe
M1 solves their three-term quadratic, e.g. $n^{2} - 3 n - 40 = 0$ , $(n - 8) (n + 5) = 0$
A1 $n = 8$ only; if $n = - 5$ (or $n = 0$ ) is also found, it must be rejected

How this is examined

Counting selections with a condition: from two groups, exactly so many, at least so many, a named person in or out: a Find about a committee or team, split into parts of one to three marks each A product such as $^{7} C_{4} \times^{5} C_{2}$ for a split between two groups (M1). For 'at least', either every allowed case listed and added, or the total minus the unwanted cases (M1). Then the correct whole number (A1). Write the expression before the value, so the method can be credited even if the arithmetic slips.
Arrangements with a restriction: fixed ends, objects together or apart, digits where 0 cannot come first: a two- or three-mark Find about the letters of a word, people in a row, or numbers formed from given digits The restricted positions counted first and multiplied by the free ones, such as $2 \times 1 \times 5!$ . A block counted as one item and multiplied by the orders inside it, such as $6! \times 2!$ . 'Apart' found as the total minus 'together'. With 0 among the digits, separate cases for the last digit, added.
Turning $^{n} C_{r}$ or $^{n} P_{r}$ into an equation in $n$ and solving it: a two-mark Show (that) a combination equals a given product in n, followed by a Find for n The factorial form written and cancelled to a product such as $\frac{n (n - 1)}{2}$ , a correct equation in $n$ , a valid solving method, and only the positive whole-number root given, with any negative root rejected in writing.
Deciding between a permutation and a combination, and saying why: the first step of every counting part, or a one-mark Explain asking why order does or does not matter A reason tied to the context, such as 'a team is the same 6 people whatever order they are chosen in, so order does not matter'. Writing only 'it is a combination' gives no reason.
Evaluating $n!$ , $^{n} P_{r}$ and $^{n} C_{r}$ without a calculator: the arithmetic inside any counting question on Paper 1 Factorials cancelled before multiplying, such as $\frac{10!}{7!} = 10 \times 9 \times 8$ , and $0! = 1$ used when $r = 0$ or $r = n$ . A count must be a whole number, so a fraction left over means a cancelling error.

Words to know

Factorial: $n! = n \times (n - 1) \times \dots \times 2 \times 1$ for a positive whole number $n$ , with $0! = 1$ . It is the number of ways to arrange $n$ different objects in a row.
Permutation: An arrangement in which order matters. $^{n} P_{r} = \frac{n!}{(n - r)!}$ is the number of ways to arrange $r$ objects taken from $n$ different objects.
Combination: A selection in which order does not matter. $^{n} C_{r} = (\binom{n}{r}) = \frac{n!}{r! (n - r)!}$ is the number of ways to choose $r$ objects from $n$ different objects.
Multiplication rule: If one choice can be made in $a$ ways and then a second choice in $b$ ways, the two together can be made in $a \times b$ ways.
Swap test: The check used in this lesson: swap two chosen objects. A different outcome means order matters (permutation); the same outcome means it does not (combination).
Block method: For objects that must be together: count them as one item, arrange the items, then multiply by the number of orders inside the block.

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