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.
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In how many ways can the 26 letters of the English alphabet be arrange [#permalink]
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Competition Mode Question In how many ways can the 26 letters of the English alphabet be arranged so that there are seven letters between the letters A and B? A. \(12\) B. \(36 * 24!\) C. \(24C7 * 18! * 2\) D. \(26C7 × 20! × 2\)E. \(26P7 × 20! × 2\)Are You Up For the Challenge: 700 Level Questions _________________
Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] In how many ways can the 26 letters of the English alphabet be arranged so that there are seven letters between the letters A and B? A. 12B. 36∗24!C. 24C7∗18!∗2D. 26C7×20!×2E. 26P7×20!×2Between A & B, there are 7 letters out of 24.Number of ways of selecting 7 letters out of 24 = 24C7Arranging these 7 numbers between A & B = 7!Number of ways of arranging A & B = 2Number of ways 9 letter word with A & B at both ends = 24C7*7!*2Consider this 9 letter word as a single letterNumber of ways of arranging the 18 letters = 18!Total number of ways the 26 letters of the English alphabet be arranged so that there are seven letters between the letters A and B = 24C7*7!*2*18! = (24!/7!17!)*7!*2*18! = 24! *2*18 = 36*24!IMO B _________________
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Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] (1) Scheme A,_,_,_,_,_,_,_,B,(remaining) A can be positioned from letters 1 to 18. Accordingly, B is positioned from 9 to 26. The remaining 24 letters can fill the spaces with no repetition allowed in 24! ways. ----> No of ways = 18*24!(2) Similarly, the reverse scheme B,_,_,_,_,_,_,_,A,(remaining) is also applicable----> Additional no of ways = 18*24!Total no of ways = 18*24! + 18*24! = 36*24!FINAL ANSWER IS (B) Posted from my mobile device
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Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] A & B have to be separated by 7 lettersMake a block of 9 letters with {A/B, _, _, _, _, _, _, _, B/A} and remaining 17 lettersWe can arrange 7 letters from 24 available in \(24p_7\) waysA & B can be interchanged in 2 ways18 units (1 block + remaining 17 letters) can be arranged in 18! ways--> Total number of ways = 2*24p7*18! = 2*{24*23*22*21*20*19*18}*18! = 2*24!*18 = 36*24! Option B
Originally posted by Dillesh4096 on 20 Mar 2020, 01:07.
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In how many ways can the 26 letters of the English alphabet be arrange [#permalink]
Quote: In how many ways can the 26 letters of the English alphabet be arranged so that there are seven letters between the letters A and B?A. 12B. 36∗24!C. 24C7∗18!∗2D. 26C7×20!×2 E. 26P7×20!×2 TOTAL 26 LETTERSA__[7]__B___[17]___The number ways to choose [7] in between is 24C7The number of arrangements of [7] is 7!The number of arrangements of A and B is 2The number of arrangements of [17] and "A_[7]_B" as a group is 18!Total arrangements 24!/7!17! * 7! * 2 * 18! = 36*24!Ans (B)
Originally posted by exc4libur on 20 Mar 2020, 03:29.
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Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] If we take out A and B from the 26 letters, we are left with 24 letters.The number of ways that 7 letters can be selected out of 24 = 24C7.The number of ways each of the 7 letters selected can be arranged = 7!After selecting 7 letters out of 24, the number of letters left = 24-7=17In how many ways can we arrange the remaining 17 letters? 17!Each of the nine letters formed by interspersing 7 letters between letters A and B can be arranged between the 17 remaining letters in 17+1=18 waysLast but not the least, the two letters A and B can be arranged interchangeably at the ends in 2 waysTotal No. of ways = \(24C7*7!*17!*18*2\) = \((24!/(7!*17!))\)\(*7!*17!*18*2\) = \(36*24!\) The answer is B.
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Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] 7 letters must separate A and B.Implication:In the 26-letter arrangement, A and B must occupy the ends of a 9-letter block, as in the following examples:A=1 __ __ __ __ __ __ __ B=9A=2 __ __ __ __ __ __ __ A=10A=17 __ __ __ __ __ __ __ B=25 A=18 __ __ __ __ __ __ __ B=26 The value in red indicates that the number of ways to position this 9-letter block = 18Number of ways to arrange A and B at the ends of this block = 2!Number of ways to arrange the remaining 24 letters = 24!To combine these options, we multiply:18 * 2! * 24! = 36 * 24! _________________
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Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] In how many ways can the 26 letters of the English alphabet be arranged so that there are seven letters between the letters A and B?A. 12B. 36∗24!C. 24C7∗18!∗2D. 26C7×20!×2E. 26P7×20!×2Assumption: Arrangements is linear not circular.A + seven alphabets + B = 9 alphabets.A and B can be written in 2 ways:1. AB 2. BARest of the places can be filled with any random alphabets that are left i.e. 24 alphabets.A_ _ _ _ _ _ _ B _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ (each space is an alphabet)_ A_ _ _ _ _ _ _ B _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ A_ _ _ _ _ _ _ B _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _......_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ A _ _ _ _ _ _ _ BSince either of A or B can take place till 9th last alphabet i.e. 18th position, total positions that A or B can take = 18So, Total ways of arrangements = 24! * 18 * 2 = 24! * 36Answer B. _________________
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Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] 26 - 9 = 17 ... If we consider the 9 letters as 1 objects we have to permute 18 different objects .. b and A can interchange ... I think the answer is C
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Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] A _ _ _ _ _ _ _ B(24C7, as A and B are fixed) + remaining letters(18!) +B _ _ _ _ _ _ _ A (24C7)+ remaining letters(18!)--------------------------------------------------=> 24C7 * 18! * 2 Ans.C
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Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] In how many ways can the 26 letters of the English alphabet be arranged so that there are seven letters between the letters A and B?The position of A, B and the other 7 letters can be fixed in the following ways :A 24C7 B B 24C7 A So, there can be 2 * 24C7 arrangements for these 9 letters. Now we can say these 9 letters are serving as a single letter. So the total letters become 18. These 18 letters can be arranged in 18! ways. So, the total arrangements will become : 2* 24C7 * 18!. C is the answer.
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Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] Arranging these 7 numbers between A & B = 24*23*.....18Number of ways of arranging A & B = 2!Considering all the 9 leeters as one arrangement Number of ways of arranging the 18 letters = 18!Therefore the tolal way of arranging the letters equal= 24*23....18*18!*2 =>24! *2*18 = 36*24 Therefore IMO B
Re: In how many ways can the 26 letters of the English alphabet be arrange [#permalink] 26 Jul 2021, 22:35 |
How many ways can two letters be selected from the English alphabet if repetition is allowed a 650?
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