AMC 10 Step by Step

Basic Counting

Multiplication principle, permutations, combinations, counting by listing

57
primary-topic problems (4.4% of all)
94
more as a secondary topic
Where it appears
20
P1-10
10
P11-15
12
P16-20
15
P21-25

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: 22 letters then 33 digits make 26⋅26⋅10⋅10⋅10=676,00026\cdot26\cdot10\cdot10\cdot10 = 676{,}000 license plates.
  • Permutations (order matters). nn different objects can be put in order in n!n! ways. Picking kk of them in order: P(n,k)=n!(n−k)!P(n,k) = \frac{n!}{(n-k)!} ways. Example: a president, vice president, and secretary from 1010 people: 10⋅9⋅8=72010\cdot9\cdot8 = 720.
  • Combinations (order does not matter). Choosing kk of nn different objects: (nk)=n!k! (n−k)!\binom{n}{k} = \frac{n!}{k!\,(n-k)!}. Also (nk)=(nn−k)\binom{n}{k} = \binom{n}{n-k} (choosing who is in is choosing who is out), and (n0)+(n1)+⋯+(nn)=2n\binom n0+\binom n1+\dots+\binom nn = 2^n (each object is in or out). Example: a 33-person committee from 1010 people: 7203!=120\frac{720}{3!} = 120, since each committee has 3!3! orders.
  • Repeated objects: n!a! b!⋯\frac{n!}{a!\,b!\cdots}, where aa, bb, … count each group of identical objects. Example: ALABAMA has four A's, so 7!4!=210\frac{7!}{4!} = 210 arrangements.
  • Complementary counting. Count what you do not want and subtract from the total. Example: 54=6255^4 = 625 four-digit numbers use only odd digits, so 9000−625=83759000-625 = 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 (nk)\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 ((n2)−n\binom{n}{2}-n: pairs of corners minus the sides), crossing points of chords ((n4)\binom{n}{4} for nn 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

  1. Decide whether order matters. Then build the object stage by stage and multiply.
  2. If the total is quick to find, count the opposite and subtract.
  3. Split into cases that do not overlap, based on the most limited spot (first digit, largest number, a special object), and add.
  4. Look for a one-to-one correspondence with something you can count directly: strings of 00s and 11s, subsets, grid paths. Example: subsets of {1,2,3,4,5}\{1,2,3,4,5\} match the 00-11 strings of length 55 (11 means "in"), so there are 25=322^5 = 32.
  5. 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) 126126 (B) 210210 (C) 330330 (D) 336336 (E) 420420

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 00 would be the smallest digit, so it would come first, which is not allowed. Choose 44 digits from 11 to 99: (94)=126\binom94 = 126.

Step 2: decreasing. Now a 00 would come last, which is fine. Choose 44 digits from 00 to 99: (104)=210\binom{10}{4} = 210. (For example, {0,3,5,9}\{0,3,5,9\} gives 95309530.)

Step 3: add. No number is both: 126+210=336126+210 = 336.

The answer is (D) 336\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 88 choices after a 11 but 11 choice after an 88.
  • Mixing up (nk)\binom{n}{k} and P(n,k)P(n,k). A committee has no order; a line-up does.
  • Forgetting that the first digit cannot be 00, while 00 is fine elsewhere. There are 900900 three-digit numbers, not 10001000.
  • Counting things twice in "at least one" problems. Example: 33-person committees from 55 boys and 55 girls with at least one girl: "pick a girl, then any 22 others" gives 5⋅(92)=1805\cdot\binom92 = 180, but the answer is (103)−(53)=110\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)

Problems, easiest first

Open any of them to try it here; the worked solution is on the problem's own page.