9 x 8 x 7 x 6 x 5
There are only 5 places on the shelf. You have 7 books to choose from. We will ignore the order of the books on the shelf. The first place can be filled from a choice of 7 books, the next place from 6, the next place from 5, the next from 4, and the last of the 5 places from 3 books. So the number of ways of choosing the 5 is found from 7 * 6 * 5 * 4 * 3 = 2520
Assuming that each of the nine books is unique, that there are only nine positions open on the given shelf, and that each book can fit in each position, the answer is 9! (nine factorial) which is equal to 362,880. If the nine books are not all unique (i.e. there are two of the same book), the number should decrease taking into effect that reversing the positions of the identical books makes no overall change. If there is one additional position, the number should increase to 10! or 3,628,800. If there is more than one additional position, the number will still increase, but not to 11! or 12! since both open spaces are reversible just like the identical books.
The letters ABC can be arranged in 3! = 3 x 2 x 1 = 6 ways. This is because there are 3 letters to arrange, and for each position, there are 3 choices for the first letter, 2 choices for the second letter, and 1 choice for the last letter. Therefore, the total number of ways to arrange the letters ABC is 6.
Situation 1: Organizing 6 itemsLet's say you are placing 6 books on a shelf, how many possible combinations of 6 items are possible?You could pick any of the six books for the first spot. For the second spot you have five books to choose from. You have four options for the third spot, three for the fourth spot, two options for the fifth spot, and the last book goes in the sixth spot.Mathematically speaking, that would be 6 x 5 x 4 x 3 x 2 x 1 = 720 possible combinations.Situation 2: Choosing # of itemsYou have finished sorting the books and you are hungry. You decide to go to McDonalds and buy a hamburger. You can pick 6 toppings, cheese, lettuce, tomatoes, mustard, ketchup, and onions. If you are only allowed two toppings, how many possible topping combinations are there? If you could choose three toppings? Four toppings? Let's break it down.If you could have one topping, you have 6combinations.If you could have two toppings, you have 30 combinations. Using the same logic as with the books, there are six toppings for the first spot and five toppings for the second spot. That is 6 x 5 = 30 options.If you could have three toppings, you have 120combinations. Six toppings for the first spot, five for the second, and three for the third topping. That is 6 x 5 x 4 = 120combinations.If you could have four toppings, you have 6 x 5 x 4 x 3 = 360 combinations.If you could have four toppings, you have 360 x 2 = 720combinations.
No, extremely high or low values will not affect the median. Because the median is the middle number of a series of numbers arranged from low to high, extreme values would only serve as the end markers of the values.
There are only 5 places on the shelf. You have 7 books to choose from. We will ignore the order of the books on the shelf. The first place can be filled from a choice of 7 books, the next place from 6, the next place from 5, the next from 4, and the last of the 5 places from 3 books. So the number of ways of choosing the 5 is found from 7 * 6 * 5 * 4 * 3 = 2520
Assuming that each of the nine books is unique, that there are only nine positions open on the given shelf, and that each book can fit in each position, the answer is 9! (nine factorial) which is equal to 362,880. If the nine books are not all unique (i.e. there are two of the same book), the number should decrease taking into effect that reversing the positions of the identical books makes no overall change. If there is one additional position, the number should increase to 10! or 3,628,800. If there is more than one additional position, the number will still increase, but not to 11! or 12! since both open spaces are reversible just like the identical books.
While massaging the question into some form that can be answered,I'm assuming:-- exactly 4 of the 10 books are math books-- the other 6 are not math books-- you're asking how many different ways all 10 can be arrangedon the shelf while keeping the math books all together.If that's your question, then the answer is as follows. If that's not your question,then the following is the answer to my question but it won't help you at all :The 10 books are really only 7 units to be arranged ... 6 random books, andone unit comprised of math books that might as well be glued together becausethey can't separated. So the shelf arrangement involves lining up 7 things.The first can be any one of the seven. For each of those . . .The second one can be any one of the remaining six. For each of those . . .The third can be any one of the remaining five. For each of those . . ...etc.Total number of possible arrangements is (7 x 6 x 5 x 4 x 3 x 2) = 5,040 ways.
Placing only US history books on a bookshelf will create a specific look. The titles of the books are less important then the size and colors of the books. Some old war history books would work well.
If the librarian follows a standard scheme, such as the Dewey Decimal, then there is essentially only one way. If the librarian arranged books randomly it would become extremely difficult to find any books: the library would become unusable and the librarian would get sacked. So again not many ways of arranging books before you stop being a librarian.
There are 39 books in the old testament only.
There were only 2 domesday books written
I only know of 2 books
There are 2. Only at the start but you can buy books for it too.
it would be 1x8x7x6x5x4x3x2=40320 Becuase if i had 5 books and they could be in any order then 5 can be in the front then 4 can be next then 3 then 2 then 1 so 5x4x3x2x1=120 But if only 1 can be in front, 8 can be next, then 7, then 6....
There is said to only be six books.
The Wolf's Rain manga only has two books.