Combinations

What a combination counts. Given $n$ distinct objects, a combination of size $r$ is any subset containing $r$ of them, with no regard to the order in which the objects were picked. The number of such subsets is written $^nC_r$ or $\binom{n}{r}$:

where: $n$ is the total number of distinct objects available, $r$ is the size of the group being chosen, and $n!$ (read "$n$ factorial") is $n \times (n-1) \times \cdots \times 1$, with $0! = 1$ by convention.

Where the formula comes from, in one line. A permutation $^nP_r$ counts every ordered arrangement of $r$ objects from $n$. Group those arrangements by which $r$ objects they use: each group of $r$ objects can be internally reordered in exactly $r!$ ways, and all of those orderings collapse to the same combination. So ordered count equals unordered count times the reorderings of each group:

where: $^nP_r = \dfrac{n!}{(n-r)!}$ is the number of ordered arrangements of $r$ objects out of $n$, already known from permutations.

Three properties worth memorising, each provable straight from the formula:

• $^nC_r = {}^nC_{n-r}$ — choosing $r$ objects to keep is equivalent to choosing the other $n-r$ objects to discard, so the two counts must agree.
• $^nC_0 = {}^nC_n = 1$ — there is exactly one way to select nothing at all, and exactly one way to select everything.
• $^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$ — fix one extra object among $n+1$; either it is left out of the chosen $r$ (the remaining $r$ come from the other $n$, giving $^nC_r$) or it is included (the other $r-1$ come from the other $n$, giving $^nC_{r-1}$). This identity generates Pascal's triangle.

The keyword to hunt for in a word problem is whether the chosen objects play distinct roles. "Select a committee," "choose a team," "pick a group of books" — none of these attach a different job to each chosen item, so they are combinations. "Elect a president and a secretary," "arrange books on a shelf," "assign first, second and third prize" — here each position is different, so they are permutations. When a problem mixes both (a group is chosen and then one member is additionally given a special role), it usually breaks into a combination step followed by a permutation or multiplication step.

Problem. From a group of 10 students, a committee of 4 is to be formed, and separately, a box holding 5 red balls and 6 blue balls needs exactly 2 red and 1 blue ball drawn for a display. Solve both.

Step 1 (committee). A committee has no internal ranking, so this is a pure combination with $n = 10$, $r = 4$:

The committee can be formed in 210 ways.

Step 2 (balls, stage one). Only the composition of the drawn balls matters, not the order of drawing, so this is again combinations, split into two independent choices. Choosing 2 red balls from 5: $^5C_2 = \dfrac{5!}{2!\,3!} = \dfrac{5\times4}{2\times1} = 10$.

Step 3 (balls, stage two and combine). Choosing 1 blue ball from 6: $^6C_1 = \dfrac{6!}{1!\,5!} = 6$. Both choices happen together to build one final selection, so by the fundamental principle of counting the counts multiply, not add: $10 \times 6 = 60$.

The final answers: the committee can be formed in 210 ways, and the ball selection can be made in 60 ways.

Competitive stretch

Problem. From the same box of 5 red and 6 blue balls, in how many ways can 3 balls be drawn such that at least 2 are red?

Step 1. "At least 2 red" out of 3 balls splits into two disjoint cases: exactly 2 red and 1 blue, or exactly 3 red and 0 blue. Since these cases cannot both happen at once, their counts are added (not multiplied).

Step 2. Exactly 2 red and 1 blue was already computed above: $^5C_2 \times {}^6C_1 = 10 \times 6 = 60$. Exactly 3 red and 0 blue: $^5C_3 \times {}^6C_0 = \dfrac{5!}{3!\,2!}\times 1 = 10 \times 1 = 10$.

Step 3. Add the disjoint cases: $60 + 10 = 70$. There are 70 ways to draw 3 balls with at least 2 red, a compact illustration of how "at least" problems in combinations decompose into an addition over the fundamental principle of counting's multiplied sub-cases.

• Using $^nC_r$ when roles are distinct. "Choose a captain and a vice-captain" is not a plain committee selection — the two posts differ, so this needs permutations, or a combination followed by an explicit ordering step.
• Adding instead of multiplying across independent stages. When a selection has several categories (red balls and blue balls above), each stage's combination count is multiplied together, since all stages happen simultaneously to form one final group; only genuinely disjoint cases (like "at least" problems) get added.
• Skipping the smaller-index shortcut. By $^nC_r = {}^nC_{n-r}$, a value like $^{10}C_8$ is faster computed as $^{10}C_2 = 45$ than by expanding $8!$ needlessly.
• Mixing up the two formulas mid-calculation — dividing by $r!$ twice, or confusing $\dfrac{n!}{r!(n-r)!}$ (combinations) with $\dfrac{n!}{(n-r)!}$ (permutations).

Ask the tutor: Not sure whether your next word problem needs $^nC_r$ or $^nP_r$? Ask the AI tutor to walk through the order-matters test on it with you.