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To arrange 3 algebra books and 3 geometry books on a shelf so that no two books of the same type are next to each other, you can use the concept of permutations. There are 3! ways to arrange the algebra books and 3! ways to arrange the geometry books. This gives a total of 3! * 3! = 36 ways to arrange the books on the shelf such that no two books of the same type are next to each other.
Ill skip the factorials and just give you what the calculator does in the end. 8*7*6*5*4*3= 20,160
6 ways
12!/(5!*7!)The number of ways to arrange nitems is n!, where "!" is the factorial function. The number of ways we can arrange the 12 books is therefore 12!. However, we don't really care what order the first 5 books are in, or what order the last 7 books are in, as long as they're the same books. We therefore divide by the number of ways to arrange 5 books and the number of ways to arrange 7 books.
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30 ways.
The answer to this one is 24. You can do this mathematically by 4*3*2*1.
The answer would be 7! or (7*6*5*4*3*2*1)=5040
5 books can be lined up on a shelf in (5 x 4 x 3 x 2 x 1) = 120 different sequences.
120. You do 5*4*3*2*1=120. you multiply the number that you are given for example how many times can books 3 be arranged on a shelf you multiply 3*2*1=6 that is your answer
Yes. They are real.
10!
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
You can wind up with 10 different pairs of books in your hand, which you can choose from a shelf of 5 books in 20 different ways.
Ill skip the factorials and just give you what the calculator does in the end. 8*7*6*5*4*3= 20,160
it depends on the shelf of course