A combination is a selection of objects where order does not matter. Choosing a committee, a hand of cards, a set of questions to attempt — in each case the group $\{A, B, C\}$ is the same group however you write it. This is the single distinction that separates combinations from permutations.
The number of combinations of $n$ distinct objects taken $r$ at a time is written ${}^{n}C_{r}$ or $\binom{n}{r}$:
$${}^{n}C_{r} = \binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}, \quad 0 \le r \le n$$
The formula is no accident. To arrange $r$ objects you could first select them ($\binom{n}{r}$ ways) and then order them ($r!$ ways), so ${}^{n}P_{r} = \binom{n}{r} \times r!$. Rearranging gives the combination formula directly — a combination is a permutation with the $r!$ internal orderings divided out.
$$\binom{n}{r} = \dfrac{{}^{n}P_{r}}{r!}$$
The clearest way to keep the two ideas apart is a side-by-side comparison:
Key properties. Several identities recur throughout the syllabus and are worth knowing by sight. The symmetry property:
$${}^{n}C_{r} = {}^{n}C_{n-r}$$
Choosing $r$ objects to keep is the same act as choosing the $n-r$ objects to leave behind, so the two counts must agree. This gives the boundary values ${}^{n}C_{0} = {}^{n}C_{n} = 1$, and a useful corollary: if ${}^{n}C_{a} = {}^{n}C_{b}$ with $a \ne b$, then $a + b = n$. The Pascal identity:
$${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$$
This is the rule behind Pascal's triangle and a stepping stone to the Binomial Theorem: it says that selections of $r$ from $n+1$ objects split cleanly into those that include a fixed object and those that exclude it.
Restricted selections. Many problems force certain objects in or out before you count:
- Always include $k$ specific objects: those $k$ are fixed, so you only choose the remaining $r-k$ from the other $n-k$ objects — ${}^{(n-k)}C_{(r-k)}$.
- Always exclude $k$ specific objects: discard them, then choose all $r$ from the surviving $n-k$ objects — ${}^{(n-k)}C_{r}$.
- "At least" / "at most" conditions: break into cases by the number taken from each group, count each case by multiplying the separate combinations, then add the cases.
A few standard applications are worth recognising on sight: the number of ways to choose $2$ points to form a line from $n$ points (no three collinear) is ${}^{n}C_{2}$; the number of diagonals of an $n$-sided polygon is ${}^{n}C_{2} - n$; and the number of triangles from $n$ points (no three collinear) is ${}^{n}C_{3}$.
Deeper Insight — "order matters" is the only question you must answer: Nearly every mistake in this chapter comes from confusing a selection with an arrangement, so train yourself to ask one thing before reaching for any formula: if I shuffle the chosen objects, is it a different outcome? If yes, it is a permutation; if no, it is a combination. A committee of three from ten is $\binom{10}{3} = 120$, but if those three become President, Secretary and Treasurer the answer jumps to ${}^{10}P_{3} = 720$ — exactly $3! = 6$ times larger, because each committee can be ranked in $6$ ways. The symmetry ${}^{n}C_{r} = {}^{n}C_{n-r}$ is more than a shortcut for arithmetic; it reflects a genuine duality between what you take and what you leave, and it is why $\binom{n}{r}$ peaks in the middle and tapers symmetrically. Master the order question and the rest of combinatorics — binomial coefficients, probability, even basic statistics — rests on solid ground.
In how many ways can a committee of 3 members be selected from a group of 8 people?
Solution- A committee has no internal ranking, so order does not matter — this is a combination.
- ${}^{8}C_{3} = \dfrac{8!}{3!\,5!} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1}$.
- $8 \times 7 \times 6 = 336$ and $3! = 6$, so $\dfrac{336}{6} = 56$.
Answer: $56$ committees.
A committee of 3 men and 2 women is to be formed from 6 men and 4 women. In how many ways can this be done?
Solution- Choose the men and the women separately, then combine the choices (the two selections happen together, so multiply).
- Men: ${}^{6}C_{3} = \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
- Women: ${}^{4}C_{2} = \dfrac{4 \times 3}{2 \times 1} = 6$.
- Total $= 20 \times 6 = 120$.
Answer: $120$ ways.
Evaluate ${}^{12}C_{10}$ using the symmetry property.
Solution- By symmetry, ${}^{n}C_{r} = {}^{n}C_{n-r}$, so ${}^{12}C_{10} = {}^{12}C_{2}$.
- ${}^{12}C_{2} = \dfrac{12 \times 11}{2 \times 1}$.
- $\dfrac{132}{2} = 66$.
Answer: ${}^{12}C_{10} = 66$.
From a standard deck of 52 cards, in how many ways can 5 cards be chosen so that all are spades? (There are 13 spades.)
Solution- The order of cards in a hand does not matter, so this is a combination from the $13$ spades.
- ${}^{13}C_{5} = \dfrac{13!}{5!\,8!} = \dfrac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1}$.
- Numerator: $13 \times 12 \times 11 \times 10 \times 9 = 154{,}440$. Denominator: $5! = 120$.
- $\dfrac{154{,}440}{120} = 1{,}287$.
Answer: $1{,}287$ five-spade hands.
Find $n$ if ${}^{n}C_{8} = {}^{n}C_{12}$.
Solution- If ${}^{n}C_{a} = {}^{n}C_{b}$ with $a \ne b$, then by symmetry the indices must satisfy $a + b = n$.
- Here $8 \ne 12$, so $n = 8 + 12$.
- $n = 20$.
Answer: $n = 20$.
A student must answer 7 questions out of 10 in an exam, but the first 3 questions are compulsory. In how many ways can the student choose the questions to attempt?
Solution- The first $3$ questions must be attempted, so they are fixed — no choice there.
- That leaves $7 - 3 = 4$ more questions to choose from the remaining $10 - 3 = 7$ questions.
- Order of selection does not matter: ${}^{7}C_{4} = {}^{7}C_{3} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}$.
- $\dfrac{210}{6} = 35$.
Answer: $35$ ways.
Verify the Pascal identity ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$ for $n = 6$, $r = 3$.
Solution- Left side: ${}^{6}C_{3} + {}^{6}C_{2}$. Now ${}^{6}C_{3} = \dfrac{6 \times 5 \times 4}{6} = 20$ and ${}^{6}C_{2} = \dfrac{6 \times 5}{2} = 15$.
- So the left side $= 20 + 15 = 35$.
- Right side: ${}^{7}C_{3} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = \dfrac{210}{6} = 35$.
- Both sides agree.
Answer: Verified: ${}^{6}C_{3} + {}^{6}C_{2} = {}^{7}C_{3} = 35$.
A committee of $5$ is to be chosen from $7$ men and $5$ women so that it contains at least $3$ women. In how many ways?
Solution- "At least $3$ women" out of $5$ committee members splits into three mutually exclusive cases: exactly $3$, $4$, or $5$ women.
- $3$W, $2$M: ${}^{5}C_{3} \times {}^{7}C_{2} = 10 \times 21 = 210$.
- $4$W, $1$M: ${}^{5}C_{4} \times {}^{7}C_{1} = 5 \times 7 = 35$.
- $5$W, $0$M: ${}^{5}C_{5} \times {}^{7}C_{0} = 1 \times 1 = 1$.
- Add the cases: $210 + 35 + 1 = 246$.
Answer: $246$ committees.
In how many ways can $11$ players be chosen from $15$ if (a) a particular player is always included, (b) a particular player is always excluded?
Solution- (a) Fix the included player; choose the remaining $11 - 1 = 10$ from the other $15 - 1 = 14$ players: ${}^{14}C_{10} = {}^{14}C_{4} = \dfrac{14 \times 13 \times 12 \times 11}{24} = 1{,}001$.
- (b) Drop the excluded player; choose all $11$ from the remaining $14$: ${}^{14}C_{11} = {}^{14}C_{3} = \dfrac{14 \times 13 \times 12}{6} = 364$.
Answer: (a) $1{,}001$ ways; (b) $364$ ways.
How many (a) diagonals and (b) triangles can be drawn using the $12$ vertices of a regular dodecagon? (No three vertices are collinear.)
Solution- (a) Joining any $2$ of the $12$ vertices gives a line segment: ${}^{12}C_{2} = \dfrac{12 \times 11}{2} = 66$. Of these, $12$ are the polygon's sides, so the diagonals number $66 - 12 = 54$.
- (b) A triangle needs any $3$ vertices, and no three are collinear, so every choice gives a genuine triangle: ${}^{12}C_{3} = \dfrac{12 \times 11 \times 10}{6} = 220$.
Answer: (a) $54$ diagonals; (b) $220$ triangles.
How many ways are there to choose a cricket team of $11$ from $17$ players that includes exactly $4$ bowlers, given $5$ of the $17$ are bowlers and the rest are non-bowlers?
Solution- Of the $17$ players, $5$ are bowlers and $17 - 5 = 12$ are non-bowlers.
- Choose exactly $4$ bowlers from the $5$: ${}^{5}C_{4} = 5$.
- The remaining $11 - 4 = 7$ players come from the $12$ non-bowlers: ${}^{12}C_{7} = {}^{12}C_{5} = \dfrac{12 \times 11 \times 10 \times 9 \times 8}{120} = 792$.
- Multiply (both selections happen together): $5 \times 792 = 3{,}960$.
Answer: $3{,}960$ teams.
A bag has $5$ red and $6$ black balls. In how many ways can $3$ red and $2$ black balls be selected, and what is the relation ${}^{n}P_{r} = r! \cdot {}^{n}C_{r}$ checked for the red selection if order mattered?
Solution- Selection ignores order, so use combinations: red ${}^{5}C_{3} = 10$ and black ${}^{6}C_{2} = 15$.
- Total selections $= 10 \times 15 = 150$.
- Check the relation for the reds: if order mattered, arrangements would be ${}^{5}P_{3} = 60$. Indeed $r! \cdot {}^{n}C_{r} = 3! \times {}^{5}C_{3} = 6 \times 10 = 60 = {}^{5}P_{3}$.
Answer: $150$ selections; and ${}^{5}P_{3} = 3! \cdot {}^{5}C_{3} = 60$ confirms ${}^{n}P_{r} = r!\cdot{}^{n}C_{r}$.