How many permutations can be made for 5 different books on a shelf that can accommodate exactly this five books?
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Madi K. More 1 Expert Answer
Gerald B. answered • 04/10/19 Experienced Statistics Tutor This is an example of a permutation, because the order of the books matters. If the order didn't matter, it would be a combination. The formula for the number of permutations of n objects taken r at a time is given by this formula: P(n,r)=n!/(n-r)! where ! is the factorial operator. In our example, n=5 and r=3, so we have: P(5,3) = 5!/(5-3)! = 5!/2! = 5*4*3*2*1/2*1 = 5*4*3 = 60 Still looking for help? Get the right answer, fast.ORFind an Online Tutor Now Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. You can divide your task in steps, count the ways that you can conduct each step and then use the multiplication principle to count the total number of ways that you can conduct your task. a) You can do it in two steps. 1.Step. Order your books. You can do it in $25!$ ways. 2.Step. Decide on how many books will be placed on each shelf. That is equivalent to writing on piece of paper 10 nonnegative (zeros are here allowed in c they are not!) integers $x_1,x_2,\dots,x_{10}$ so that they add up to 25. For $i=1,2,\ldots,10$, $x_i$ represents the number of books that will be placed on the $i$-th shelf. The total number of ways that you can conduct this step is equal to the number of solutions to the equation $$x_1+x_2+\ldots+x_{10}=25,$$ with $0\leq x_i, i=1,2,\ldots,10$. This is a typical stars and bars problem with $n=10$ and $k=25$. Then (according to the Wikipedia page, Stars and Bars, Theorem two) you can do that in $$\dbinom{n+k-1}{n-1}=\dbinom{10+25-1}{10-1}=\dbinom{34}{9}$$ So in sum the task in (a) can be conducted in $25!\times\dbinom{34}{9}$ ways. c). Based on a) the answer is straightforward by allowing in Step 2. only positive values for the $x_i, i=1,2,\ldots,10$. The total number of ways that you can conduct this step is equal to the number of solutions to the equation $$x_1+x_2+\ldots+x_{10}=25,$$ with $1\leq x_i, i=1,2,\ldots,10$. Then (according to the Wikipedia page, Stars and Bars, Theorem one) you can do it in $\dbinom{k-1}{n-1}=\dbinom{25-1}{10-1}$ ways. So in sum the task in (c) can be conducted in $25!\times\dbinom{24}{9}$ ways. b) 25C25 is equal to 1 so, your approach here is not correct. Here you can think it as follows. Write on each book a number from 1 to 10 (this number indicates the shelf that it will be placed). So you can do it $10\times10\times\ldots\times10=10^{25}$ ways. (With steps as in the previous questions: 1.Step Choose a shelf for book Nr1. You can do it in 10 ways. 2.Step Choose a shelf for book Nr2. You can do it in 10 ways and so on until 25.Step Choose a shelf for book Nr25. You can do it in 10 ways. So by multiplication rule you can do it in $10^{25}$ ways.) $\begingroup$ I am solving number of arrangements for following question: Eight books are placed on a shelf. Three of them form a 3-volume series, two form a 2-volume series, and 3 stand on their own. In how many ways can the eight books be arranged so that the books in the 3-volume series are placed together according to their correct order, and so are the books in the 2-volume series? Noted that there is only one correct order for each series. I analyzed this problem in following way: 3 volume book as A, 2 volume book as B, and remaining books as C. So three can be arranged as 3!, and C in turn can be arranged as 3! which by product rule we can have 3! * 3! = 36. Why this analysis or approach to this problem is wrong. Correct answer is 120 (i.e. 5!). Kindly help. thanks asked Sep 17, 2020 at 9:53
$\endgroup$ 1 $\begingroup$ You can think of the three books who make a volume as a single big book (their relative positions can't change) and so for the other two volumes books, they can be considered as a single one too since their relative position can't change, therefore you have the equivalent of 5 books: (3 books), (2 books), (1 book), (1 book), (1 book) And the number of ways you can arrange them is $5!$ indeed. answered Sep 17, 2020 at 10:00
Baffo rastaBaffo rasta 3633 silver badges15 bronze badges $\endgroup$ $\begingroup$ Since the $3$-volume series are placed together according to their correct order, we have $A_{1}A_{2}A_{3}$ and for the $2$ volume series we also have $B_{1}B_{2}.$ So we can consider the $3$ volume books to be a single book denoted by $A$ and the $2$ volume books to be a single book denoted by $B$. Thus we need to arrange $A,B$ and the remaining $3$ books. This is the same as the number of ways to rearrange $5$ books. Since the number of ways in which $5$ books can be arranged is $5!=120,$ we have $120$ ways. answered Sep 17, 2020 at 10:06
Alessio KAlessio K 10.5k9 gold badges15 silver badges31 bronze badges $\endgroup$ 2 How many permutations can be made for 5 different books on a shelf that can accommodate only 4 books?There are 5 possible shelves the first book can be placed on. For each of those, there are then 4 possible shelves the second book can be placed on. 5 * 4 = 20.
How many ways can 5 different books be arranged on a shelf if I there are no restrictions II 2 books are always together III 2 books are never together?Solution, The number of ways by which books can be arranged on the shelf is 120, 48, and 72 respectively.
How many ways can 5 different books be arranged on a shelf if two specified books must not be side by side?× 3 × 2 ! = 72.
How many ways can you arrange 5 things?As illustrated before for 5 objects, the number of ways to pick from 5 objects is 5! . Note that your choice of 5 objects can take any order whatsoever, because your choice each time can be any of the remaining objects. So we say that there are 5 factorial = 5! = 5x4x3x2x1 = 120 ways to arrange five objects.
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