A permutation counts ordered arrangements. Many questions only care which objects are chosen: a committee, a hand of cards, pizza toppings. Then we count each group once.
Key idea
The number of ways to choose $r$ objects from $n$ different objects, when order does not matter, is
$$\binom{n}{r} = {_nC_r} = \dfrac{_nP_r}{r!} = \dfrac{n!}{r!\,(n-r)!}.$$
Why it works
Count the ordered choices first: $_nP_r$. Each unordered group of $r$ objects shows up in that count once for every way of ordering it, which is $r!$ times. Dividing by $r!$ leaves each group counted exactly once. Choosing which $r$ to take is the same as choosing which $n - r$ to leave, so $\binom{n}{r} = \binom{n}{n-r}$.
Worked example: A committee
In how many ways can a committee of 3 be chosen from 10 people?
- Ordered choices: $_{10}P_3 = 10 \cdot 9 \cdot 8 = 720$.
- Each committee appears in $3! = 6$ orders.
- $\binom{10}{3} = \dfrac{720}{6} = 120$.
Answer: 120
Worked example: Two groups at once
Choose 2 juniors from 6 and 3 seniors from 5. How many ways?
- Juniors: $\binom{6}{2} = 15$. Seniors: $\binom{5}{3} = 10$.
- Every junior pair goes with every senior trio: $15 \times 10 = 150$.
Answer: 150
Worked example: At least one
A class of 10 includes 3 chess players. How many teams of 4 include at least one chess player?
- All teams: $\binom{10}{4} = 210$.
- Teams with no chess players (choose from the other 7): $\binom{7}{4} = 35$.
- At least one = all − none: $210 - 35 = 175$.
Answer: 175
Common mistake
Deciding "order matters" from the wording "first, second..." alone. Ask what makes two outcomes different. Three officers with different jobs: order matters. Three people getting identical prizes: it does not. Check by swapping two chosen people: a new outcome means permutation; the same outcome means combination.
The numbers $\binom{n}{r}$ are the entries of Pascal's triangle and the coefficients in the binomial theorem. They will reappear in the binomial distribution.