See also: probability

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The calculator below will calculate number of permutations or combinations.

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Permutation is an ordered arrangement of a number of elements of a set.

Mathematically, given a set with $n$ numbers of elements, the number of permutations of size $r$ is denoted by $P(n,r)$ or ${}_{n}P_r$ or ${}^{n}P_r$.

The formula is given by

$$P(n,r)={}_{n}P_r={}^{n}P_r=\frac{n!}{(n-r)!}$$

where $n!$ ($n$ factorial) $=n\times (n-1)\times (n-2)\times \mathrm{...}\times 1$ and $0!=1$.

For example, given the set of letters $\{\text{a},\text{b},\text{c}\}$, the permutations of size 2 (take 2 elements of the set) are $\{\text{a},\text{b}\}$, $\{\text{b},\text{a}\}$, $\{\text{a},\text{c}\}$, $\{\text{c},\text{a}\}$, $\{\text{b},\text{c}\}$, and $\{\text{c},\text{b}\}$. Please note that the order is important; $\{\text{a},\text{b}\}$ is considered different from $\{\text{b},\text{a}\}$.

The number of permutations is 6.

$$\begin{array}{rl}P(3,2)={}_{3}P_2={}^{3}P_2& =\frac{3!}{(3-2)!}\\ & =\frac{3\times 2\times 1}{1!}\\ & =\frac{6}{1}\\ & =6\end{array}$$

Another example: How many different ways are there can 5 different books be arranged on the self?

Answer: Here, $n=5$ and $r=5$.
So,

$$\begin{array}{rl}{}^{5}P_5& =\frac{5!}{(5-5)!}\\ & =\frac{5\times 4\times 3\times 2\times 1}{0!}\\ & =\frac{120}{1}\\ & =120\end{array}$$

As can be seen from the above example, when $n=r$, then ${}^{n}P_r=n!$.

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Combination is an unordered arrangement of a number of elements of a set.

Given a set with $n$ numbers of elements, the number of combinations of size $r$ is denoted by $C(n,r)$ or ${}_{n}C_r$ or ${}^{n}C_r$.

The formula is given by

$$C(n,r)={}_{n}C_r={}^{n}C_r=\frac{n!}{r!(n-r)!}$$

For example, given the set of letters $\{\text{a},\text{b},\text{c}\}$, the combinations of size 2 (take 2 elements of the set) are $\{\text{a},\text{b}\}$, $\{\text{a},\text{c}\}$, and $\{\text{b},\text{c}\}$. Please note that the order is not important; $\{\text{b},\text{a}\}$ is considered the same as $\{\text{a},\text{b}\}$.

The number of combinations is 3.

$$\begin{array}{rl}C(3,2)={}_{3}C_2={}^{3}C_2& =\frac{3!}{2!(3-2)!}\\ & =\frac{3\times 2\times 1}{2\times 1\times 1!}\\ & =\frac{6}{2}\\ & =3\end{array}$$

Another example: A basket contains an apple, an orange, a pear, and a banana. How many combinations of three fruits are there?

Answer: Here, $n=4$ and $r=3$.
So,

$$\begin{array}{rl}{}^{4}C_3& =\frac{4!}{3!(4-3)!}\\ & =\frac{4\times 3\times 2\times 1}{(3\times 2\times 1)1!}\\ & =\frac{24}{6}\\ & =4\end{array}$$

For combination, when $n=r$, the number of combinations is always equal to 1.

By Jimmy Sie

See also: probability