Permutations and combinations: counting with and without order

2026-10-22 huynhvietkhanh Comments: 0

Distinguish ordered selections from subsets, derive the counting formulas, and count selections containing exactly a specified number of one type.

High schoolCounting & Probability

Content updated: October 22, 2026

Both permutations and combinations select elements from a finite set, but permutations also record their order. The key question is therefore whether interchanging two selected elements produces a new outcome.

Learning goals: distinguish ordered from unordered selections, choose the appropriate formula, and count selections involving two types of objects.

Prerequisites: the addition and multiplication principles (English lesson) and permutations (English lesson).

On this page: The idea · Method · Examples · Practice · Further learning

1. Selection without repetition: does order matter?

Let a set contain $n$ distinct elements, and let $k$ be an integer with $0\le k\le n$. A combination of size $k$ is a subset containing $k$ elements: order does not matter, and no element is repeated. A $k$-permutation is an ordered list of $k$ pairwise distinct elements selected from the set.

Their numbers are denoted by $C_n^k$ and $A_n^k$, respectively. When $k=0$, there is one empty subset and one empty list, so both numbers equal $1$. When $k=n$, an ordered selection is a permutation of the entire set.

2. Counting formulas and two identities for combinations

Selecting the entries of an ordered list in succession leaves $n,n-1,\ldots,n-k+1$ choices. By the multiplication principle,

$$A_n^k=n(n-1)\cdots(n-k+1)=\frac{n!}{(n-k)!}.$$

Each subset with $k$ elements can be ordered in exactly $k!$ ways. Forgetting the order in a permutation produces a combination, and each combination has been counted exactly $k!$ times. Hence

$$A_n^k=k!C_n^k,\qquad C_n^k=\frac{n!}{k!(n-k)!}.$$

Thus the number of ordered selections is $k!$ times the number of unordered selections; the two numbers agree when $k=0$ or $k=1$. The symmetry identity, valid for $0\le k\le n$, is

$$C_n^k=C_n^{n-k}.$$

Indeed, $C_n^{n-k}=n!/((n-k)!k!)=C_n^k$. Combinatorially, choosing $k$ elements is equivalent to specifying the $n-k$ elements to omit. Pascal’s identity, valid for $0\le k\le n-1$, is

$$C_n^k+C_n^{k+1}=C_{n+1}^{k+1}.$$

Taking a common denominator and using $(k+1)+(n-k)=n+1$, we obtain

$$\begin{aligned}C_n^k+C_n^{k+1}&=\frac{n!(k+1)+n!(n-k)}{(k+1)!(n-k)!}\\&=\frac{(n+1)!}{(k+1)!(n-k)!}=C_{n+1}^{k+1}.\end{aligned}$$

Alternatively, partition the subsets of size $k+1$ in a set of size $n+1$ according to whether they contain a fixed element. The two classes contain $C_n^k$ and $C_n^{k+1}$ subsets, respectively.

3. Examples and selection problems

Example 1. Choosing three students for a work activity

A class has $10$ students. Choose $3$ of them for a work activity without assigning roles. Changing the order of their names does not change the selected group, so the number of choices is

$$C_{10}^3=\frac{10!}{3!7!}=120.$$

For example, $ABC,ACB,BAC,BCA,CAB,CBA$ all describe the same group of three students $A,B,C$. These six lists must not be counted as six groups.

Example 2. Choosing students and assigning three offices

In the same class of $10$ students, choose three different students as class president, vice president, and secretary, with one office per person. Listing their names in that fixed order of offices gives

$$A_{10}^3=10\cdot9\cdot8=720.$$

For three fixed students $A,B,C$, the six lists above represent six different assignments of offices. The result $720=120\cdot3!$ can also be obtained by first choosing the group and then assigning the offices.

General problem. Selecting exactly a specified number of one type

A batch contains $N$ distinct products: $N_A$ of type $A$ and $N-N_A$ of other types. Select an unordered group of $n$ products without replacement, containing exactly $k$ products of type $A$. Choosing $k$ products of type $A$ can be done in $C_{N_A}^k$ ways; for each such choice, there are $C_{N-N_A}^{n-k}$ ways to select the rest. The number of groups is

$$C_{N_A}^kC_{N-N_A}^{n-k}.$$

Here the parameters are integers satisfying $0\le N_A\le N$ and $0\le n\le N$. Such a selection is possible precisely when

$$\max\{0,n-(N-N_A)\}\le k\le\min\{n,N_A\}.$$

Outside this range there are $0$ choices; do not substitute negative integers into factorials. The word “random” does not by itself require a probability calculation: this problem counts groups, and each product is a distinct object even when several products have the same type.

Example 3. Two red balls among three selected balls

A box contains $10$ balls: $6$ red and $4$ white. Select $3$ distinct balls simultaneously, with exactly $2$ red balls. Choosing two red balls and then one white ball gives

$$C_6^2C_4^1=15\cdot4=60.$$

Do not multiply by $3!$, because the order of drawing is not part of the outcome being counted. Counting ordered sequences of three draws would be a different problem.

4. Practice exercises

From $8$ students, choose an unordered group of $2$. Then consider choosing a class president and a different vice president. How many choices are there in each case?
Using the box in Example 3, select $3$ balls containing exactly $1$ red ball. How many groups are possible?
Use Pascal’s identity to evaluate $C_7^3+C_7^4$.

Show answers and hints

Unordered groups: $C_8^2=28$. Two distinct offices: $A_8^2=8\cdot7=56$.
$C_6^1C_4^2=6\cdot6=36$.
$C_7^3+C_7^4=C_8^4=70$.

5. Common mistakes

Using permutations for a group with no assigned roles, or combinations for distinct offices.
Allowing repeated elements while applying a formula for selection without repetition.
Multiplying by an extra factorial when order is not counted.
Ignoring the available number of each type, or taking the factorial of a negative integer.

6. Related lessons

Original exercise resource on Scribd.