In this section you will learn to In this section we will address the following two problems.
The first problem comes under the category of Circular Permutations, and the second under Permutations with Similar Elements.
Suppose we have three people named A, B, and C. We have already determined that they can be seated in a straight line in 3! or 6 ways. Our next problem is to see how many ways these people can be seated in a circle. We draw a diagram.  It happens that there are only two ways we can seat three people in a circle, relative to each other’s positions. This kind of permutation is called a circular permutation. In such cases, no matter where the first person sits, the permutation is not affected. Each person can shift as many places as they like, and the permutation will not be changed. We are interested in the position of each person in relation to the others. Imagine the people on a merry-go-round; the rotation of the permutation does not generate a new permutation. So in circular permutations, the first person is considered a place holder, and where he sits does not matter.
The number of permutations of \(n\) elements in a circle is \((n-1)!\)
In how many different ways can five people be seated at a circular table? Solution We have already determined that the first person is just a place holder. Therefore, there is only one choice for the first spot. We have So the answer is 24.
In how many ways can four couples be seated at a round table if the men and women want to sit alternately? Solution We again emphasize that the first person can sit anywhere without affecting the permutation. So there is only one choice for the first spot. Suppose a man sat down first. The chair next to it must belong to a woman, and there are 4 choices. The next chair belongs to a man, so there are three choices and so on. We list the choices below. So the answer is 144.
Let us determine the number of distinguishable permutations of the letters ELEMENT. Suppose we make all the letters different by labeling the letters as follows. \[E_1LE_2ME_3NT \nonumber \] Since all the letters are now different, there are 7! different permutations. Let us now look at one such permutation, say \[LE_1ME_2NE_3T \nonumber \] Suppose we form new permutations from this arrangement by only moving the E's. Clearly, there are 3! or 6 such arrangements. We list them below. \begin{aligned} &\mathrm{LE}_{1} \mathrm{ME}_{2} \mathrm{NE}_{3} \\ &\mathrm{LE}_{1} \mathrm{ME}_{3} \mathrm{NE}_{2} \\ &\mathrm{LE}_{2} \mathrm{ME}_{1} \mathrm{NE}_{3} \mathrm{T} \\ &\mathrm{LE}_{2} \mathrm{ME}_{3} \mathrm{NE}_{1} \mathrm{T} \\ &\mathrm{LE}_{3} \mathrm{ME}_{2} \mathrm{NE}_{1} \mathrm{T} \\ & \mathrm{LE}_{3} \mathrm{ME}_{I} \mathrm{NE}_{2} \mathrm{T} \end{aligned} Because the E's are not different, there is only one arrangement LEMENET and not six. This is true for every permutation. Let us suppose there are n different permutations of the letters ELEMENT. Then there are \(n \cdot 3!\) permutations of the letters \(E_1LE_2ME_3NT\). But we know there are 7! permutations of the letters \(E_1LE_2ME_3NT\). Therefore, \(n \cdot 3! = 7!\) Or \(n = \frac{7!}{3!}\). This gives us the method we are looking for.
The number of permutations of n elements taken \(n\) at a time, with \(r_1\) elements of one kind, \(r_2\) elements of another kind, and so on, is \[\frac{n !}{r_{1} ! r_{2} ! \ldots r_{k} !} \nonumber \]
Find the number of different permutations of the letters of the word MISSISSIPPI. Solution The word MISSISSIPPI has 11 letters. If the letters were all different there would have been 11! different permutations. But MISSISSIPPI has 4 S's, 4 I's, and 2 P's that are alike. So the answer is \(\frac{11!}{4!4!2!} = 34,650\).
If a coin is tossed six times, how many different outcomes consisting of 4 heads and 2 tails are there? Solution Again, we have permutations with similar elements. We are looking for permutations for the letters HHHHTT. The answer is \(\frac{6!}{4!2!} = 15\).
In how many different ways can 4 nickels, 3 dimes, and 2 quarters be arranged in a row? Solution Assuming that all nickels are similar, all dimes are similar, and all quarters are similar, we have permutations with similar elements. Therefore, the answer is \[\frac{9 !}{4 ! 3 ! 2 !}=1260 \nonumber \]
A stock broker wants to assign 20 new clients equally to 4 of its salespeople. In how many different ways can this be done? Solution This means that each sales person gets 5 clients. The problem can be thought of as an ordered partitions problem. In that case, using the formula we get \[\frac{20 !}{5 ! 5 ! 5 ! 5 !}=11,732,745,024 \nonumber \]
A shopping mall has a straight row of 5 flagpoles at its main entrance plaza. It has 3 identical green flags and 2 identical yellow flags. How many distinct arrangements of flags on the flagpoles are possible? Solution The problem can be thought of as distinct permutations of the letters GGGYY; that is arrangements of 5 letters, where 3 letters are similar, and the remaining 2 letters are similar: \[ \frac{5 !}{3 ! 2 !} = 10 \nonumber \] Just to provide a little more insight into the solution, we list all 10 distinct permutations: GGGYY, GGYGY, GGYYG, GYGGY, GYGYG, GYYGG, YGGGY, YGGYG, YGYGY, YYGGG We summarize.
1. Circular Permutations The number of permutations of n elements in a circle is \[(n -1)! \nonumber \] 2. Permutations with Similar Elements The number of permutations of n elements taken n at a time, with \(r_1\) elements of one kind, \(r_2\) elements of another kind, and so on, such that \(\mathrm{n}=\mathrm{r}_{1}+\mathrm{r}_{2}+\ldots+\mathrm{r}_{\mathrm{k}}\) is \[\frac{n !}{r_{1} ! r_{2} ! \dots r_{k} !} \nonumber \] This is also referred to as ordered partitions.
Method 1: We begin by arranging three blue balls and three red balls in a row. There are $\binom{6}{3}$ such arrangements since we must choose which three of the six spaces will be occupied by the blue balls. This creates seven spaces in which we can place three green balls, five spaces between successive balls and two at the ends of the row. For instance, $$\square \color{red}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{red}{\bullet} \square \color{blue}{\bullet} \square \color{red}{\bullet} \square$$ We now wish to insert three green balls so that no two of them are adjacent. To do so, we must choose three of these seven spaces, which can be done in $\binom{7}{3}$ ways. One such arrangement is balls and two at the ends of the row. For instance, $$\color{green}{\bullet} \color{red}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{red}{\bullet} \color{blue}{\bullet} \color{red}{\bullet}$$ The green balls represent the positions that can be occupied by Alvin, John, and Albert; the blue balls represent the positions of the other three people; the red balls represent the positions of the three unoccupied seats. We can arrange Alvin, John, and Albert in the places occupied by green balls in $3!$ ways. We can arrange the remaining three people in the positions occupied by the blue balls in $3!$ ways. Consequently, the number of permissible seating arrangements is $$\binom{6}{3}\binom{7}{3}3!3!$$ Method 2: We use the Inclusion-Exclusion Principle. There are $\binom{9}{3}$ ways of choosing the locations of the empty seats and $6!$ ways of arranging the six people in the remaining seats. From these seating arrangements, we must exclude those arrangements in which the three students who do not wish to sit in adjacent seats sit in adjacent seats. If two of Alvin, John, and Albert sit together, we have eight objects to arrange, the pair of seats in which we will place those students, the three empty seats, and the other four students. There are $\binom{8}{3}$ ways to choose three of the eight positions for the empty seats, $\binom{3}{2}$ ways to choose two of the three students to sit together, $5!$ ways to arrange the remaining objects, and $2!$ ways to arrange the pair of chosen students in the designated pair of seats. Subtracting this from the total removes those cases in which all three students sit together twice, once when we designate the leftmost two students as the pair and once when we designate the rightmost two students as the pair. Since we only wish to exclude these arrangements once, we must add them back. If all three students sit together, we have seven objects to arrange, the three empty seats, the trio of seats occupied by Alvin, John, and Albert, and the other three people. We can select three of these seven positions for the empty seats in $\binom{7}{3}$ ways. We can arrange the remaining four objects in $4!$ ways. We can arrange Alvin, John, and Albert within the designated trio of seats in $3!$ ways. Hence, the number of seating arrangements in which Alvin, John, and Albert sit together is $$\binom{7}{3}4!3!$$ By the Inclusion-Exclusion Principle, the number of permissible seating arrangements is $$\binom{9}{3}6! - \binom{8}{3}\binom{3}{2}5!2! + \binom{7}{3}4!3!$$ |
In how many ways can 6 students be seated in a row if 2 particular students are to be separated
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