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Comprehensive Guide on Permutations

Probability and Statistics
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Probability Theory
schedule Oct 31, 2022
Last updated
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Let's start with a motivating example before we state the formal definition of permutations.

Example.

Total number of ways to arrange letters

Suppose we have $7$ letters: $\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$, $\mathrm{D}$, $\mathrm{E}$, $\mathrm{F}$ and $\mathrm{G}$. We select $3$ letters. How many different arrangements are there?

Solution. The word "arrangement" means that the ordering matters. For instance, $\mathrm{ABC}$ is considered to be different from $\mathrm{BAC}$. Let's illustrate the scenario:

We have $3$ available slots, and we have to fill each slot with a different letter.

  • for the 1st slot, we have 7 letters to choose from.

  • for the 2nd slot, since we have already picked a letter for the 1st slot, we only have 6 letters to choose from.

  • for the 3rd slot, since two letters have already been chosen, we have 5 letters left to choose from.

The total number of ways to fill the slots is equal to the product of the number of ways to fill each slot:

$$\begin{equation}\label{eq:D3iQ7UTUTFCzzCLKSH6} 7\times6\times5=210 \end{equation}$$

Let's now generalize the calculation by defining the following variables:

  • $n$ is the total number of elements to initially choose from.

  • $r$ is the number of elements to choose.

In our example, $n=7$ since we have $7$ letters, and $r=3$ because we must choose $3$ letters. Now recall how we calculated the total number of arrangements in \eqref{eq:D3iQ7UTUTFCzzCLKSH6}. Using $n$ and $r$, this number can be computed by:

$$\begin{equation}\label{eq:ioMaMIui2M87EGPWm5o} \frac{n!}{(n-r)!} \end{equation}$$

To understand why this works, let's substitute $n=7$ and $r=3$ as in our example:

$$\frac{7!}{(7-3)!}= \frac{7!}{4!}=7\times6\times5$$

Notice how this is exactly how we computed the total number of arrangements in \eqref{eq:D3iQ7UTUTFCzzCLKSH6}. Mathematicians prefer to use the word "permutations" instead of "arrangements" to emphasize the fact that ordering matters.

Definition.

Definition and formula for permutation

Permutation refers to the number of ways of ordering $r$ elements from a total of $n$ elements, and is computed by:

$$_nP_r = \frac{n!}{(n-r)!}$$

Where:

  • $n$ is the total number of elements.

  • $r$ is the number of elements to select.

Let's now go through some examples of applying this formula.

Example.

Constructing numbers from digits

Suppose we have six digits from $1$ to $6$. How many 3-digit numbers can we construct from these digits if each digit can only be used once?

Solution. Permutation applies in this case because the $3$-digit number $456$ is considered different from $546$, and hence the ordering matters. Let $n=6$ and $r=3$ because we're selecting $3$ digits out of a total of $6$ digits. Applying the formula for permutation gives:

$$\begin{align*} _6P_3&=\frac{6!}{(6-3)!}\\ &=\frac{6!}{3!}\\ &= 6\times5\times4\\ &= 120 \end{align*}$$

Therefore, a total of $120$ digits can be constructed.

Example.

Rearranging balls

If we had $5$ balls numbered from $1$ to $5$, how many different ways could we order exactly two of them?

Solution. Once again, permutation applies because we are concerned with the ordering of the balls. Let $n=5$ and $r=2$ because we have $5$ elements, from which we select $2$ elements. We now apply the formula for permutation:

$$\begin{align*} _5P_2&=\frac{5!}{(5-2)!}\\ &=\frac{5!}{3!}\\ &=5\times4\\ &=20 \end{align*}$$

Therefore, there are $20$ ways of ordering $2$ out of $5$ balls.

robocat
Published by Isshin Inada
Edited by 0 others
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