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Basic Counting

Multiplication principle, permutations, combinations, counting by listing

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Basic CountingAMC 10 Step by Step

Name: Date: October 28, 2026

20 problems, easiest first

What you need to know

Multiplication principle. If a job has stages, and each stage has the same number of choices no matter what you chose before, multiply. Cases that do not overlap add. Example: $2$ letters then $3$ digits make $26\cdot26\cdot10\cdot10\cdot10 = 676{,}000$ license plates.
Permutations (order matters). $n$ different objects can be put in order in $n!$ ways. Picking $k$ of them in order: $P(n,k) = \frac{n!}{(n-k)!}$ ways. Example: a president, vice president, and secretary from $10$ people: $10\cdot9\cdot8 = 720$ .
Combinations (order does not matter). Choosing $k$ of $n$ different objects: $\binom{n}{k} = \frac{n!}{k!\,(n-k)!}$ . Also $\binom{n}{k} = \binom{n}{n-k}$ (choosing who is in is choosing who is out), and $\binom n0+\binom n1+\dots+\binom nn = 2^n$ (each object is in or out). Example: a $3$ -person committee from $10$ people: $\frac{720}{3!} = 120$ , since each committee has $3!$ orders.
Repeated objects: $\frac{n!}{a!\,b!\cdots}$ , where $a$ , $b$ , … count each group of identical objects. Example: ALABAMA has four A's, so $\frac{7!}{4!} = 210$ arrangements.
Complementary counting. Count what you do not want and subtract from the total. Example: $5^4 = 625$ four-digit numbers use only odd digits, so $9000-625 = 8375$ have an even digit.

How AMC 10 tests it

Problems 3–10: "How many three-digit numbers / license plates / subsets have property X?", by multiplying or by one $\binom{n}{k}$ .
Problems 10–18: choose-then-arrange questions (committees, handshakes, seating) where the trap is whether order matters.
Digit problems that come down to short casework on the first digit.
Counting the right thing: pairs of people, diagonals ( $\binom{n}{2}-n$ : pairs of corners minus the sides), crossing points of chords ( $\binom{n}{4}$ for $n$ points on a circle, if no three chords meet at one point).
The count is often the hidden step inside a probability problem.

Standard approaches

Decide whether order matters. Then build the object stage by stage and multiply.
If the total is quick to find, count the opposite and subtract.
Split into cases that do not overlap, based on the most limited spot (first digit, largest number, a special object), and add.
Look for a one-to-one correspondence with something you can count directly: strings of $0$ s and $1$ s, subsets, grid paths. Example: subsets of $\{1,2,3,4,5\}$ match the $0$ - $1$ strings of length $5$ ( $1$ means "in"), so there are $2^5 = 32$ .
Check with a tiny case you can list by hand.

Worked example

How many four-digit positive integers have digits that are strictly increasing from left to right or strictly decreasing from left to right?

(A) $126$ (B) $210$ (C) $330$ (D) $336$ (E) $420$

Solution. Any four different digits can be written in increasing order in exactly one way, and in decreasing order in one way. So we count sets of four digits.

Step 1: increasing. A $0$ would be the smallest digit, so it would come first, which is not allowed. Choose $4$ digits from $1$ to $9$ : $\binom94 = 126$ .

Step 2: decreasing. Now a $0$ would come last, which is fine. Choose $4$ digits from $0$ to $9$ : $\binom{10}{4} = 210$ . (For example, $\{0,3,5,9\}$ gives $9530$ .)

Step 3: add. No number is both: $126+210 = 336$ .

The answer is $\boxed{\textbf{(D)}\ 336}$ .

Pitfalls

Multiplying when one stage's choices depend on an earlier choice. Use casework. Example: in a two-digit number whose second digit is bigger than the first, the second digit has $8$ choices after a $1$ but $1$ choice after an $8$ .
Mixing up $\binom{n}{k}$ and $P(n,k)$ . A committee has no order; a line-up does.
Forgetting that the first digit cannot be $0$ , while $0$ is fine elsewhere. There are $900$ three-digit numbers, not $1000$ .
Counting things twice in "at least one" problems. Example: $3$ -person committees from $5$ boys and $5$ girls with at least one girl: "pick a girl, then any $2$ others" gives $5\cdot\binom92 = 180$ , but the answer is $\binom{10}{3}-\binom53 = 110$ . Complementary counting avoids this.

Traps that recur

Summing binom(9,k) 2^k instead of binom(9,k) 2^(k-1), which gives 3^9 - 1 = 19682, choice (E); the factor 2^(k-1) is right because the largest digit's position is forced. (2025 AMC 10A #24)
Multiplying the 81 multiples of 11 by the number of permutations, which double counts triples like {2, 0, 9} that give two different multiples (209 and 902). (2017 AMC 10A #25)
Answering 70 by assuming no three diagonals are concurrent; the regular octagon has concurrency at the center and at eight other points. (2013 AMC 10A #25)
Forgetting that the two-friends case includes two separate triangles as well as one six-person loop, or overcounting loops by ignoring rotations and reflections. (2012 AMC 10A #23)

1.2024 AMC 10B #1
In a long line of people arranged left to right, the 1013th person from the left is also the 1010th person from the right. How many people are in the line?
1. A) $2021$
2. B) $2022$
3. C) $2023$
4. D) $2024$
5. E) $2025$
2.2018 AMC 10B #3
In the expression $\left(\underline{\qquad}\times\underline{\qquad}\right)+\left(\underline{\qquad}\times\underline{\qquad}\right)$ each blank is to be filled in with one of the digits $1,2,3,$ or $4,$ with each digit being used once. How many different values can be obtained?
1. A) $2$
2. B) $3$
3. C) $4$
4. D) $6$
5. E) $24$
3.2018 AMC 10B #5
How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
1. A) $128$
2. B) $192$
3. C) $224$
4. D) $240$
5. E) $256$
4.2018 AMC 10B #1
Kate bakes a $20$ -inch by $18$ -inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?
1. A) $90$
2. B) $100$
3. C) $180$
4. D) $200$
5. E) $360$
5.2008 AMC 10B #1
A basketball player made $5$ baskets during a game. Each basket was worth either $2$ or $3$ points. How many different numbers could represent the total points scored by the player?
1. A) $2$
2. B) $3$
3. C) $4$
4. D) $5$
5. E) $6$
6.2004 AMC 10B #2
How many two-digit positive integers have at least one $7$ as a digit?
1. A) $10$
2. B) $18$
3. C) $19$
4. D) $20$
5. E) $30$
7.2004 AMC 10B #1
Each row of the Misty Moon Amphitheater has $33$ seats. Rows $12$ through $22$ are reserved for a youth club. How many seats are reserved for this club?
1. A) $297$
2. B) $330$
3. C) $363$
4. D) $396$
5. E) $726$
8.2024 AMC 10A #9
In how many ways can $6$ juniors and $6$ seniors form $3$ disjoint teams of $4$ people so that each team has $2$ juniors and $2$ seniors?
1. A) $720$
2. B) $1350$
3. C) $2700$
4. D) $3280$
5. E) $8100$
9.2023 AMC 10A #6
An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube?
1. A) $42$
2. B) $63$
3. C) $84$
4. D) $126$
5. E) $252$
10.2022 AMC 10B #3
How many three-digit positive integers have an odd number of even digits?
1. A) $150$
2. B) $250$
3. C) $350$
4. D) $450$
5. E) $550$
11.2020 AMC 10A #6
How many $4$ -digit positive integers (that is, integers between $1000$ and $9999$ , inclusive) having only even digits are divisible by $5?$
1. A) $80$
2. B) $100$
3. C) $125$
4. D) $200$
5. E) $500$
12.2020 AMC 10B #5
How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
1. A) $210$
2. B) $420$
3. C) $630$
4. D) $840$
5. E) $1050$
13.2017 AMC 10A #8
At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?
1. A) $240$
2. B) $245$
3. C) $290$
4. D) $480$
5. E) $490$
14.2015 AMC 10A #3
Ann made a 3-step staircase using 18 toothpicks as shown in the figure. How many toothpicks does she need to add to complete a 5-step staircase?
1. A) $9$
2. B) $18$
3. C) $20$
4. D) $22$
5. E) $24$
15.2013 AMC 10A #11
A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
1. A) $10$
2. B) $12$
3. C) $15$
4. D) $18$
5. E) $25$
16.2013 AMC 10A #7
A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
1. A) $6$
2. B) $8$
3. C) $9$
4. D) $12$
5. E) $16$
17.2011 AMC 10A #13
How many even integers are there between 200 and 700 whose digits are all different and come from the set {1, 2, 5, 7, 8, 9}?
1. A) $12$
2. B) $20$
3. C) $72$
4. D) $120$
5. E) $200$
18.2009 AMC 10B #11
How many $7$ -digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$ , $2$ , $3$ , $3$ , $5$ , $5$ , $5$ ?
1. A) $6$
2. B) $12$
3. C) $24$
4. D) $36$
5. E) $48$
19.2009 AMC 10B #7
By inserting parentheses, it is possible to give the expression $2\times3 + 4\times5$ several values. How many different values can be obtained?
1. A) $2$
2. B) $3$
3. C) $4$
4. D) $5$
5. E) $6$
20.2007 AMC 10B #8
On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form $bbcac,$ where $0 \le a < b < c \le 9,$ and $b$ was the average of $a$ and $c.$ How many different five-digit numbers satisfy all these properties?
1. A) $12$
2. B) $16$
3. C) $18$
4. D) $20$
5. E) $25$

Basic Counting — answersAMC 10 Step by Step

Name: Date: October 28, 2026

Keep this page separate from the problems.

Answers

#	Answer	The idea it turns on
1	B	The person in the middle is counted once from each end. Count the people on each side of that person, then add that person back once.
2	B	The value only depends on which two digits get multiplied together. There are just three ways to split 1, 2, 3, 4 into two pairs.
3	D	Count all subsets, then take away the ones with no prime at all, which are the subsets made only from 4, 6, 8, 9.
4	A	Count how many 2-inch pieces fit along each side of the pan, 10 along one side and 9 along the other, then multiply.
5	E	Each time a 2-point basket becomes a 3-point basket, the total goes up by exactly 1. Six choices for the number of 3-pointers give six totals.
6	B	It is easier to count the two-digit numbers with no 7 at all (8 choices for the tens digit, 9 for the ones digit) and take them away from 90.
7	C	Rows 12 through 22 means both end rows count, so there are 11 rows, not 10. Then multiply 11 rows by 33 seats.
8	B	Split the juniors into three pairs (15 ways), split the seniors into three pairs (15 ways), then match each junior pair with a senior pair (6 ways).
9	D	Follow one vertex's number: 3 edges use it and each edge sits on 2 faces, so it is counted 6 times. The cube's value is 6 times 21.
10	D	Whatever the first two digits are, exactly 5 of the 10 possible last digits make the count of even digits odd. So exactly half of all 900 three-digit numbers work.
11	B	The last digit must be 0 (even and divisible by 5). The first digit has 4 choices, since it cannot be 0, and each middle digit has 5.
12	B	Place the colors one group at a time: choose 3 of the 7 spots for yellow, then 2 of the remaining 4 for green, then order brown and purple.
13	B	A pair shakes hands exactly when at least one of the two is one of the 10 strangers. Count stranger-with-group pairs, then stranger-with-stranger pairs.
14	D	Count the flat toothpicks row by row, then use the fact that there are just as many standing ones. The 3-step staircase has 18, the 5-step one has 40.
15	A	Ten ways to pick 2 people means the council has 5 members. Then count the ways to pick 3 of those 5, which is also 10.
16	C	English is always in, so pick 3 more from the other 5 courses. Of those 10 choices, only History, Art, Latin has no math, so 9 are allowed.
17	A	The first digit must be 2 or 5 and the last digit must be 2 or 8. If the first digit is 2, the last digit can only be 8.
18	A	The middle digit has no partner, so it must be the 5 that appears an odd number of times. The first three digits are 2, 3, 5 in any order.
19	C	Only the single plus sign can be moved earlier. It can join the 3 with the left product, the right product, both, or neither, giving 26, 46, 50 and 70.
20	D	Once a and c are picked, b is their average, so it is fixed. It is a whole number exactly when a and c are both even or both odd.

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.