Permutations and Combinations is the habit of choosing whether order matters and applying factorial, binomial, and arrangement counts. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
Use permutations $P(n,k)$ when order matters and combinations $\binom{n}{k}$ when it does not.
Why It Works
1
To arrange $k$ of $n$ distinct items in order, there are $n$ choices for the first slot, $n-1$ for the second, down to $n-k+1$ for the $k$-th: that product is $\tfrac{n!}{(n-k)!} = P(n,k)$.
2
Each unordered selection of $k$ items can be arranged in $k!$ orders, so it is counted $k!$ times among the permutations.
3
Dividing out those repeats gives the count of unordered selections: $\binom{n}{k} = \dfrac{P(n,k)}{k!} = \dfrac{n!}{k!\,(n-k)!}$.
4
So the only question to ask is: does rearranging the chosen items give a genuinely different outcome? If yes, permutations; if no, combinations.
Worked Examples
Example 1
From $10$ students, how many ways are there to (a) line up $3$ for a photo, and (b) pick a committee of $3$?
1
Lining up is ordered, so use permutations: $P(10, 3) = 10 \cdot 9 \cdot 8 = 720$.
2
A committee is unordered, so use combinations: $\binom{10}{3} = \dfrac{10 \cdot 9 \cdot 8}{3!} = \dfrac{720}{6} = 120$.
3
Note the photo lineups are exactly $3! = 6$ times the committees, since each committee can be arranged $6$ ways.
Answer:
(a) $720$ lineups; (b) $120$ committees.
Warm-up
How many distinct arrangements are there of the letters in the word $\text{BALLOON}$?
1
$\text{BALLOON}$ has $7$ letters with repeats: two $L$'s and two $O$'s (and single $B, A, N$).
2
Arrangements of a multiset use the formula $\dfrac{n!}{n_1!\,n_2!\cdots}$, dividing out rearrangements of identical letters.
3
Here $\dfrac{7!}{2!\,2!} = \dfrac{5040}{4} = 1260$.
Answer:
$1260$
Contest level
A committee of $4$ is chosen from $5$ men and $4$ women. How many committees have exactly $2$ men and $2$ women?
1
Order within the committee does not matter, so use combinations and the multiplication principle.
2
Choose $2$ of the $5$ men: $\binom{5}{2} = 10$. Choose $2$ of the $4$ women: $\binom{4}{2} = 6$.
3
Multiply the independent choices: $10 \cdot 6 = 60$.
Answer:
$60$
Olympiad / Challenge
In how many ways can $8$ distinct people be seated around a round table, where seatings that are rotations of each other are considered the same?
1
There are $8!$ ways to seat $8$ people in labeled chairs.
2
Around a circle, each distinct seating is counted once for each of the $8$ rotations of the table, so we divide by $8$.
(If reflections were also identified, divide by an additional factor of $2$ to get $\tfrac{7!}{2} = 2520$, but here only rotations are identified.)
Answer:
$5040$
Going Deeper
These two formulas generalize in three key directions: arrangements of a multiset $\dfrac{n!}{n_1!\cdots n_r!}$ (divide out identical-item swaps), circular permutations $\dfrac{n!}{n}=(n-1)!$ (divide out rotations), and the symmetry $\binom{n}{k} = \binom{n}{n-k}$. Choosing then arranging ('choose a set, then order it') recovers $P(n,k) = \binom{n}{k} \cdot k!$.
Order-matters-or-not is the first decision in almost every counting problem, so this distinction underlies essentially all of competition combinatorics: lineups vs. committees, sequences vs. sets, anagrams, lottery counts, and seating problems. AMC/AIME problems often nest combinations inside the multiplication principle (choose by category, then multiply).
Pitfall: identical objects and symmetric seatings must be divided out exactly once. Treating repeated letters as distinct overcounts; forgetting to divide by $n$ for a round table overcounts by the rotation factor; and double-dividing (e.g. dividing by both rotation and reflection when only rotations are identified) undercounts. Decide precisely which arrangements are 'the same' before dividing.
Spot the Signal
Look for problems where the key step is choosing whether order matters and applying factorial, binomial, and arrangement counts.
You can describe the hard part as choosing whether order matters and applying factorial, binomial, and arrangement counts, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the counting object that calls for permutations and combinations, then rewrite the givens around it.
Name the relation that makes Permutations and Combinations legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Permutations and Combinations just because the surface looks familiar; verify the required condition first.
Applying Permutations and Combinations because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let permutations and combinations reveal the counting structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is choosing whether order matters and applying factorial, binomial, and arrangement counts.