5.2.2 Permutations

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,
nPr^{n}P_{r}This reads as nn permute rr, where nn represents the number of choices and rr 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 66, and we are to select 44 letters, then,
6P4^{6}P_{4}Which would simplify to,
360360Therefore, the final answer would be,
360360


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 AA appears twice, divide the permutation by 2!2!.

Let’s look at some past paper questions.

1. Find the number of different ways that 55 boys and 66 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,
b  G1  b  G2  b  G3  b  G4  b  G5  b  G6  b\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 b\textcolor{#2192ff}{b_{}} represent the slots in which the boys can be placed to ensure that no boy is standing next to another boy. GG represents Girl.

To get the number of different arrangements, we have to consider that there are 55 boys and only 77 slots for boys, therefore,
7P5^{7}P_{5}We also have to consider the ways in which the girls can be arranged. Since there are six girls,
6!6!Putting everything together,
7P5×6!^{7}P_{5} \times 6!Which simplifies to give,
2 520×7202\ 520 \times 7201 814 4001\ 814\ 400Therefore, the final answer is,
1 814 4001\ 814\ 4002. Find the number of different ways in which all 99 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][I, E, E, A]Now let’s try to visualize what the given problem would look like,
v  M  v  N  v  C  v  M  v  T  v\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 v\textcolor{#2192ff}{v_{}} represent the slots in which the vowels can be placed to ensure that no vowel is next to another vowel.
[I,E,E,A][I, \textcolor{#0f0}{E}, \textcolor{#0f0}{E}, A]To get the number of different arrangements, we have to consider that there are 44 vowels, and there are 66 slots for the vowels,
6P4^{6}P_{4}One of the vowels, E\textcolor{#0f0}{E}, appears twice, so we need to divide by 2!2!,
6P42!\frac{^{6}P_{4}}{\textcolor{#0f0}{2!}}We also need to consider the arrangements for the consonants,
[M,N,C,M,T][\textcolor{red}{M}, N, C, \textcolor{red}{M}, T]There are 55 consonants and M\textcolor{red}{M} appears twice, hence,
5!2!\frac{5!}{\textcolor{red}{2!}}Putting everything together,
5!2!×6P42!\frac{5!}{2!} \times \frac{^{6}P_{4}}{2!}Which simplifies to give,
60×18060 \times 18010 80010\ 800Therefore, the final answer is,
10 80010\ 8003. A security code consists of 22 letters followed by a 44-digit number. The letters are chosen from {A,B,C,D,E}\{A, B, C, D, E\} and the digits are chosen from {1,2,3,4,5,6,7}\{1, 2, 3, 4, 5, 6, 7\}. No letter or digit may appear more than once. An example of a code is BE3216BE3216. (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 55 letters and only 22 slots form them, this would be,
5P2^{5}P_{2}For the numbers, we have 77 of them and only 44 slots for them, this would be,
7P4^{7}P_{4}Putting everything together,
5P2×7P4^{5}P_{2} \times ^{7}P_{4}Which simplifies to give,
20×84020 \times 84016 80016\ 800Therefore, the final answer is,
16 80016\ 800(b) Find the number of different codes that include the letter AA or the digit 55 or both.

To solve this problem, there are three scenarios we have to consider,
Arrangements with A and no 5\textcolor{#2192ff}{\textmd{Arrangements with }A\textmd{ and no }5}Arrangements with 5 and no A\textcolor{#0f0}{\textmd{Arrangements with }5\textmd{ and no }A}Arrangements with both A and 5\textcolor{red}{\textmd{Arrangements with both }A\textmd{ and }5}Let’s evaluate the first scenario, arrangements with AA and no 55. It would look something like this,
{B,C,D,E}    {1,2,3,4,6,7}\{B, C, D, E\} \ \ \ \ \{1, 2, 3, 4, 6, 7\}A a   a a a a\underline{A}\ \underline{\textcolor{transparent}{a}}\ \ \ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}Since AA 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 55, to do that we remove 55 from the pool of selection.

Now let’s find the number of arrangements with AA and no 55,
4P1×6P4^{4}P_{1} \times ^{6}P_{4}Remember that of the two slots for the letters, AA can either be in the first slot (A a\underline{A}\ \underline{\textcolor{transparent}{a}}), or the second slot (a A\underline{\textcolor{transparent}{a}}\ \underline{A}), so we need to multiply our answer by 22,
4P1×6P4×2^{4}P_{1} \times ^{6}P_{4} \times 2Which simplifies to give,
4×360×24 \times 360 \times 22 880\textcolor{#2192ff}{2\ 880}Now let’s find the number of arrangements with 55 and no AA. It would look something like this,
{B,C,D,E}    {1,2,3,4,6,7}\{B, C, D, E\} \ \ \ \ \{1, 2, 3, 4, 6, 7\}a a   5 5 a a\underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}\ \ \ \underline{5}\ \underline{\textcolor{transparent}{5}}\ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}Since 55 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 AA, to do that we remove AA from the pool of selection.

Now let’s find the number of arrangements with 55 and no AA,
4P2×6P3^{4}P_{2} \times ^{6}P_{3}Remember that 55 can be on any of the four slots for the numbers, so we need to multiply by 44,
4P2×6P3×4^{4}P_{2} \times ^{6}P_{3} \times 4Which simplifies to give,
12×120×412 \times 120 \times 45 760\textcolor{#0f0}{5\ 760}Now let’s find the number of arrangements with both AA and 55. It would look something like this,
{B,C,D,E}    {1,2,3,4,6,7}\{B, C, D, E\} \ \ \ \ \{1, 2, 3, 4, 6, 7\}A a   5 5 a a\underline{A}\ \underline{\textcolor{transparent}{a}}\ \ \ \underline{5}\ \underline{\textcolor{transparent}{5}}\ \underline{\textcolor{transparent}{a}}\ \underline{\textcolor{transparent}{a}}Since both AA and 55 are guaranteed selections, they are removed from the pool of selection.

So the number of arrangements would be,
4P1×6P3^{4}P_{1} \times ^{6}P_{3}AA can be in any of the 22 slots for the letters, so we multiply by 22,
4P1×6P3×2^{4}P_{1} \times ^{6}P_{3} \times 255 can be in any of the 44 slots for the numbers, so we multiply by 44,
4P1×6P3×2×4^{4}P_{1} \times ^{6}P_{3} \times 2 \times 4Which simplifies to give,
4×120×2×44 \times 120 \times 2 \times 43 840\textcolor{red}{3\ 840}Adding all the three scenarios together, we get,
Total =2 880+5 760+3 840\textmd{Total } = \textcolor{#2192ff}{2\ 880} + \textcolor{#0f0}{5\ 760} + \textcolor{red}{3\ 840}Total =12 480\textmd{Total } = 12\ 480Therefore, the final answer is,
12 48012\ 480