5.2.1 Arrangements
In this topic we will learn how to
If objects are arranged in a line, we can determine the total number of different possible arrangements. We could do it manually by considering all possible arrangements but that would be tedious. So instead we will use n!, which is read as ‘n factorial’, where n represents the number of objects to be arranged.
Considering that there are no repetitions or restrictions, if we have 5 objects arranged in a line, then the total number of different arrangements possible would be 5!. We can simplify this using a calculator. Therefore our final answer would be,
5! = 120In a case where there are repetitions we would have to cater for those repetitions.
Let’s look at an example.
Example 1
Find the total number of different arrangements of the 8 letters in the word
\textmd{N\textcolor{#2192ff}{EE}DL\textcolor{#2192ff}{E}\textcolor{#0f0}{SS}}Using the idea above, our answer would be,
8!However, since the letters, E, and S are repeated, we have to divide by the factorial of the number of times the letter appears. For example, if the letter appears twice, divide by 2!, if the letter appears thrice, divide by 3!, etc,
\frac{8!}{\textcolor{#2192ff}{3!} \times \textcolor{#0f0}{2!}}If you punch that into the calculator, you get,
3\ 360Therefore, the final answer would be,
3\ 360
In some cases, the question will give you certain restrictions. We will look at how to deal with restrictions in the questions below.
Let’s look at some past paper questions.
1. (a) Find the total number of different arrangements of the 11 letters in the word CATERPILLAR. (9709/52/F/M/21 number 6)
\textmd{C\textcolor{#2192ff}{A}TE\textcolor{#0f0}{R}PI\textcolor{red}{LL}\textcolor{#2192ff}{A}\textcolor{#0f0}{R}}Since the word has 11 letters, our first thought would be to say,
11!Now let’s consider any repetitions. The letter R appears twice, so we will divide by 2!,
\frac{11!}{\textcolor{#0f0}{2!}}The letter A appears twice, so we will divide by 2!,
\frac{11!}{\textcolor{#0f0}{2!}\textcolor{#2192ff}{2!}}The letter L appears twice, so we will divide by 2!,
\frac{11!}{\textcolor{#0f0}{2!}\textcolor{#2192ff}{2!}\textcolor{red}{2!}}This simplifies to,
4\ 989\ 600Therefore, the final answer is,
4\ 989\ 600(b) Find the total number of different arrangements of the 11 letters in the word CATERPILLAR in which there is an R at the beginning and an R at the end, and the two As are not together.
In this question, you will notice that we have been restricted to keep both Rs at the ends and to keep the As apart from each other.
To find the total number of different arrangements starting and ending with an R and two As are not together, we have to consider the number of different arrangements starting and ending with an R but the two As are together then subtract that from the total number of different arrangements starting and ending with an R,
\textmd{Total of R at ends and As not together = Total of R at ends }-\textmd{ Total of R at ends and As together}
Let’s find the total number of arrangements when the Rs are at the ends and there are no other restrictions. It would look something like this,
\underline{\textcolor{#0f0}{R}}\ \underline{C}\ \ \underline{A}\ \ \underline{T}\ \ \underline{E}\ \ \underline{P}\ \ \underline{I}\ \ \underline{L}\ \ \underline{L}\ \ \underline{A}\ \underline{\textcolor{#0f0}{R}}Since the Rs have fixed positions we can ignore them in the calculation,
\underline{C}\ \ \underline{\textcolor{#2192ff}{A}}\ \ \underline{T}\ \ \underline{E}\ \ \underline{P}\ \ \underline{I}\ \ \underline{\textcolor{red}{L}}\ \ \underline{\textcolor{red}{L}}\ \ \underline{\textcolor{#2192ff}{A}}Now to find the total number of arrangements when the Rs are at the ends and there are no other restrictions, we say,
\frac{9!}{\textcolor{#2192ff}{2!}\textcolor{red}{2!}}Note: 9! because there are a total of 9 letters.
Therefore, the total number of different arrangements when the Rs are at the ends and there are no other restrictions is,
90\ 720Now let’s find the total number of different arrangements when the Rs are at the ends and the As are together. That would look something like,
\underline{\textcolor{#0f0}{R}}\ \underline{C}\ \ \underline{\textcolor{#2192ff}{AA}}\ \ \underline{T}\ \ \underline{E}\ \ \underline{P}\ \ \underline{I}\ \ \underline{L}\ \ \underline{L}\ \underline{\textcolor{#0f0}{R}}Since the As are together, we merge their blocks and they become 1. Let’s get rid of the Rs since they have fixed positions,
\underline{C}\ \ \underline{\textcolor{#2192ff}{AA}}\ \ \underline{T}\ \ \underline{E}\ \ \underline{P}\ \ \underline{I}\ \ \underline{\textcolor{red}{L}}\ \ \underline{\textcolor{red}{L}}Now to find the total number of different arrangements when the Rs are at the ends and the As are together, we say,
\frac{8!}{\textcolor{red}{2!}}Note: Count the number of underlines to get the total number of letters. If the joint letters where different, for example, \underline{BC}, you would multiply your final answer by 2 because the joint letter can be written as \underline{BC} or \underline{CB}.
Therefore, the total number of different arrangements when the Rs are at the ends and the As are together is,
20\ 160Now let’s go back to the statement we deduced above,
\textmd{Total of R at ends and As not together = Total of R at ends }-\textmd{ Total of R at ends and As together}
Let’s substitute in our answers,
\textmd{Total of R at ends and As not together } = 90\ 720 - 20\ 160Simplify,
\textmd{Total of R at ends and As not together } = 70\ 560Therefore, the final answer is,
70\ 5602. (a) Find the number of the different arrangements of the 8 letters in the word DECEIVED in which all three Es are together and the two Ds are together. (9709/51/M/J/22 number 1)
The question sets two restrictions. The first being that all three Es are together and the second being that the two Ds are together. It would look something like this,
\underline{\textcolor{#2192ff}{EEE}}\ \ \underline{\textcolor{#0f0}{DD}}\ \ \underline{C}\ \ \underline{I}\ \ \underline{V}Since there are five letters, counting by the underlines, this would give us,
5!Which simplifies to,
120Therefore, the final answer is,
120(b) Find the number of different arrangements of the 8 letters in the word DECEIVED in which the three Es are not all together.
To get the number of arrangements in which the three Es are not all together, we will subtract the number of arrangements in which the three Es are together from the total number of different arrangements,
\textmd{Total arrangements in which three Es are not all together = Total number of different arrangements }-\textmd{ Total arrangements in which Es are all together}
Let’s start by finding the total number of different arrangements,
\underline{\textcolor{#0f0}{D}}\ \ \underline{\textcolor{#2192ff}{E}}\ \ \underline{C}\ \ \underline{\textcolor{#2192ff}{E}}\ \ \underline{I}\ \ \underline{V}\ \ \underline{\textcolor{#2192ff}{E}}\ \ \underline{\textcolor{#0f0}{D}}Since there are 8 letters, the letter \textcolor{#2192ff}{E} appears three times, the letter \textcolor{#0f0}{D} appears twice, then,
\frac{8!}{\textcolor{#2192ff}{3!}\textcolor{#0f0}{2!}}This would simplify to,
3\ 360Now let’s find the total number of arrangements when the three Es are all together. It would look something like this,
\underline{\textcolor{#2192ff}{EEE}}\ \ \underline{\textcolor{#0f0}{D}}\ \ \underline{C}\ \ \underline{I}\ \ \underline{V}\ \ \underline{\textcolor{#0f0}{D}}So our number of arrangements would be,
\frac{6!}{\textcolor{#0f0}{2!}}This would simplify to,
360Now let’s go back to the statement we deduced above,
\textmd{Total arrangements in which three Es are not all together = Total number of different arrangements }-\textmd{ Total arrangements in which Es are all together}
Let’s substitute in our answers,
\textmd{Total arrangements in which three Es are not all together } = 3\ 360 - 360Simplify,
\textmd{Total arrangements in which three Es are not all together } = 3\ 000Therefore, the final answer is,
3\ 0003. Raman and Sanjay are members of a quiz team which has 9 members in total. Two photographs of the quiz team are to be taken. For the first photograph, the 9 members will stand in a line. (9709/51/O/N/21 number 5)
(a) How many different arrangements of the 9 members are possible in which Raman will be at the center of the line?
Let’s start by assigning Raman a letter of R and Sanjay, S. Having Raman in the middle would look something like this,
\underline{P_{1}}\ \ \underline{P_{2}}\ \ \underline{P_{3}}\ \ \underline{P_{4}}\ \ \underline{\textcolor{#2192ff}{R_{}}}\ \ \underline{P_{5}}\ \ \underline{P_{6}}\ \ \underline{P_{7}}\ \ \underline{S}Note: P represents person, so P_{1} would be person 1.
Since the R will be fixed we can ignore it,
\underline{P_{1}}\ \ \underline{P_{2}}\ \ \underline{P_{3}}\ \ \underline{P_{4}}\ \ \underline{P_{5}}\ \ \underline{P_{6}}\ \ \underline{P_{7}}\ \ \underline{S}Now let’s find the total number of arrangements possible for the above. There are 8 under lines and no repeats, so it would be,
8!Which simplifies to,
40\ 320Therefore, the final answer is,
40\ 320(b) How many different arrangements of the 9 members are possible in which Raman and Sanjay are not next to each other.
To find the arrangements in which Raman and Sanjay are not next to each other, we will subtract the number of arrangements in which they are next to each other from the total number of arrangements,
\textmd{Total arrangements where }R\textmd{ and }S\textmd{ are not next to each other = Total number of arrangements }-\textmd{ Total number of arrangements where }R\textmd{ and }S\textmd{ are together}
Let’s find the number of total arrangements. It would look like this,
\underline{P_{1}}\ \ \underline{P_{2}}\ \ \underline{P_{3}}\ \ \underline{P_{4}}\ \ \underline{R_{}}\ \ \underline{P_{5}}\ \ \underline{P_{6}}\ \ \underline{P_{7}}\ \ \underline{S}Since there are 9 under lines and no repetitions, this would be,
9!Which simplifies to,
362\ 880Now let’s find the number of arrangements when R and S are together. It would look something like this,
\underline{P_{1}}\ \ \underline{P_{2}}\ \ \underline{P_{3}}\ \ \underline{P_{4}}\ \ \underline{\textcolor{#2192ff}{RS_{}}}\ \ \underline{P_{5}}\ \ \underline{P_{6}}\ \ \underline{P_{7}}Since there are 8 underlines and no repetition, this would be,
8!However, note that the bracket containing \underline{\textcolor{#2192ff}{RS}} can also be written as \underline{\textcolor{#2192ff}{SR}}. For that reason we will multiply our answer by 2,
8! \times 2Which simplifies to,
80\ 640Now let’s go back to the statement we deduced above,
\textmd{Total arrangements where }R\textmd{ and }S\textmd{ are not next to each other = Total number of arrangements }-\textmd{ Total number of arrangements where }R\textmd{ and }S\textmd{ are together}
Let’s substitute in our answers,
\textmd{Total arrangements where }R\textmd{ and }S\textmd{ are not next to each other } = 362\ 880 - 80\ 640Simplify,
\textmd{Total arrangements where }R\textmd{ and }S\textmd{ are not next to each other } = 282\ 240Therefore, the final answer is,
282\ 240