5.2.2 Permutations - GCEALevelMaths9709

In this topic we will learn how to:

understand and solve simple problems involving permutations

Permutation refers to the number of different arrangements of objects where the order matters. According, to this definition, you will notice that the arrangements covered in a previous paper follow this definition. They are a special type of permutation, in which the number of choices is equal to the number of selections. In this paper, however, we will focus on permutations in which the number of choices differs from the number of selections. The notation for a permutation is as follows,
$^{n}P_{r}$ This reads as $n$ permute $r$ , where $n$ represents the number of choices and $r$ the number of selections to be made. This may appear on your calculator as, nPr.
Let’s look at an example.

Example 1
Four letters are selected from [A, B, C, D, F, E], and a letter cannot be repeated. Find the total number of different arrangements of the selected four letters.

Since the total number of choices we have is $6$ , and we are to select $4$ letters, then,
$^{6}P_{4}$ Which would simplify to,
$360$ Therefore, the final answer would be,
$360$

When dealing with permutations we can also have repetitions just like when dealing with typical arrangements. To deal with repetitions divide by the factorial of the number of times the object appears. For example, if a letter $A$ appears twice, divide the permutation by $2!$ .

Let’s look at some past paper questions.

1. Find the number of different ways that $5$ boys and $6$ girls can stand in a row if no boy stands next to another boy. (9709/62/O/N/18 number 4)

Let’s try to visualize what that would look like,
$\underline{\textcolor{#2192ff}{b_{}}}\ \ \underline{G_{1}}\ \ \underline{\textcolor{#2192ff}{b_{}}}\ \ \underline{G_{2}}\ \ \underline{\textcolor{#2192ff}{b_{}}}\ \ \underline{G_{3}}\ \ \underline{\textcolor{#2192ff}{b_{}}}\ \ \underline{G_{4}}\ \ \underline{\textcolor{#2192ff}{b_{}}}\ \ \underline{G_{5}}\ \ \underline{\textcolor{#2192ff}{b_{}}}\ \ \underline{G_{6}}\ \ \underline{\textcolor{#2192ff}{b_{}}}$ Note: The $\textcolor{#2192ff}{b_{}}$ represent the slots in which the boys can be placed to ensure that no boy is standing next to another boy. $G$ represents Girl.

To get the number of different arrangements, we have to consider that there are $5$ boys and only $7$ slots for boys, therefore,
$^{7}P_{5}$ We also have to consider the ways in which the girls can be arranged. Since there are six girls,
$6!$ Putting everything together,
$^{7}P_{5} \times 6!$ Which simplifies to give,
$2\ 520 \times 720$ $1\ 814\ 400$ Therefore, the final answer is,
$1\ 814\ 400$ 2. Find the number of different ways in which all $9$ letters of the word MINCEMEAT can be arranged when no vowel (A, E, I are vowels) is next to another vowel. (9709/61/M/J/18 number 7)

From the word MINCEMEAT, we have the vowels,
$[I, E, E, A]$ Now let’s try to visualize what the given problem would look like,
$\underline{\textcolor{#2192ff}{v_{}}}\ \ \underline{M}\ \ \underline{\textcolor{#2192ff}{v_{}}}\ \ \underline{N}\ \ \underline{\textcolor{#2192ff}{v_{}}}\ \ \underline{C}\ \ \underline{\textcolor{#2192ff}{v_{}}}\ \ \underline{M}\ \ \underline{\textcolor{#2192ff}{v_{}}}\ \ \underline{T}\ \ \underline{\textcolor{#2192ff}{v_{}}}$ Note: The $\textcolor{#2192ff}{v_{}}$ represent the slots in which the vowels can be placed to ensure that no vowel is next to another vowel.
$[I, \textcolor{#0f0}{E}, \textcolor{#0f0}{E}, A]$ To get the number of different arrangements, we have to consider that there are $4$ vowels, and there are $6$ slots for the vowels,
$^{6}P_{4}$ One of the vowels, $\textcolor{#0f0}{E}$ , appears twice, so we need to divide by $2!$ ,
$\frac{^{6}P_{4}}{\textcolor{#0f0}{2!}}$ We also need to consider the arrangements for the consonants,
$[\textcolor{red}{M}, N, C, \textcolor{red}{M}, T]$ There are $5$ consonants and $\textcolor{red}{M}$ appears twice, hence,
$\frac{5!}{\textcolor{red}{2!}}$ Putting everything together,
$\frac{5!}{2!} \times \frac{^{6}P_{4}}{2!}$ Which simplifies to give,
$60 \times 180$ $10\ 800$ Therefore, the final answer is,
$10\ 800$ 3. A security code consists of $2$ letters followed by a $4$ -digit number. The letters are chosen from $\{A, B, C, D, E\}$ and the digits are chosen from $\{1, 2, 3, 4, 5, 6, 7\}$ . No letter or digit may appear more than once. An example of a code is $BE3216$ . (9709/53/O/N/21 number 5)

(a) How many different codes can be formed?

So let’s start with the letters. So we have $5$ letters and only $2$ slots form them, this would be,
$^{5}P_{2}$ For the numbers, we have $7$ of them and only $4$ slots for them, this would be,
$^{7}P_{4}$ Putting everything together,
$^{5}P_{2} \times ^{7}P_{4}$ Which simplifies to give,
$20 \times 840$ $16\ 800$ Therefore, the final answer is,
$16\ 800$ (b) Find the number of different codes that include the letter $A$ or the digit $5$ or both.

To solve this problem, there are three scenarios we have to consider,
$\textcolor{#2192ff}{\textmd{Arrangements with }A\textmd{ and no }5}$ $\textcolor{#0f0}{\textmd{Arrangements with }5\textmd{ and no }A}$ $\textcolor{red}{\textmd{Arrangements with both }A\textmd{ and }5}$ Let’s evaluate the first scenario, arrangements with $A$ and no $5$ . It would look something like this,
$\{B, C, D, E\} \ \ \ \ \{1, 2, 3, 4, 6, 7\}$ $\underline{A}\ \underline{\textcolor{transparent}{a}}\ \ \ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}$ Since $A$ is a guaranteed selection, we remove it from the pool of selection and place it on one of the slots. This means there are four letters left in the pool of selection and only one slot for the letters. For the numbers, we want to ensure that we do not select $5$ , to do that we remove $5$ from the pool of selection.

Now let’s find the number of arrangements with $A$ and no $5$ ,
$^{4}P_{1} \times ^{6}P_{4}$ Remember that of the two slots for the letters, $A$ can either be in the first slot ( $\underline{A}\ \underline{\textcolor{transparent}{a}}$ ), or the second slot ( $\underline{\textcolor{transparent}{a}}\ \underline{A}$ ), so we need to multiply our answer by $2$ ,
$^{4}P_{1} \times ^{6}P_{4} \times 2$ Which simplifies to give,
$4 \times 360 \times 2$ $\textcolor{#2192ff}{2\ 880}$ Now let’s find the number of arrangements with $5$ and no $A$ . It would look something like this,
$\{B, C, D, E\} \ \ \ \ \{1, 2, 3, 4, 6, 7\}$ $\underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}\ \ \ \underline{5}\ \underline{\textcolor{transparent}{5}}\ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}$ Since $5$ is a guaranteed selection, we remove it from the pool of selection and place it on one of the slots. This means there are six numbers in the pool of selection and only three slots for the numbers. For the letters, we want to ensure that we do not select $A$ , to do that we remove $A$ from the pool of selection.

Now let’s find the number of arrangements with $5$ and no $A$ ,
$^{4}P_{2} \times ^{6}P_{3}$ Remember that $5$ can be on any of the four slots for the numbers, so we need to multiply by $4$ ,
$^{4}P_{2} \times ^{6}P_{3} \times 4$ Which simplifies to give,
$12 \times 120 \times 4$ $\textcolor{#0f0}{5\ 760}$ Now let’s find the number of arrangements with both $A$ and $5$ . It would look something like this,
$\{B, C, D, E\} \ \ \ \ \{1, 2, 3, 4, 6, 7\}$ $\underline{A}\ \underline{\textcolor{transparent}{a}}\ \ \ \underline{5}\ \underline{\textcolor{transparent}{5}}\ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}$ Since both $A$ and $5$ are guaranteed selections, they are removed from the pool of selection.

So the number of arrangements would be,
$^{4}P_{1} \times ^{6}P_{3}$ $A$ can be in any of the $2$ slots for the letters, so we multiply by $2$ ,
$^{4}P_{1} \times ^{6}P_{3} \times 2$ $5$ can be in any of the $4$ slots for the numbers, so we multiply by $4$ ,
$^{4}P_{1} \times ^{6}P_{3} \times 2 \times 4$ Which simplifies to give,
$4 \times 120 \times 2 \times 4$ $\textcolor{red}{3\ 840}$ Adding all the three scenarios together, we get,
$\textmd{Total } = \textcolor{#2192ff}{2\ 880} + \textcolor{#0f0}{5\ 760} + \textcolor{red}{3\ 840}$ $\textmd{Total } = 12\ 480$ Therefore, the final answer is,
$12\ 480$