it depends on the shelf of course
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.
5! = 5 x 4 x 3 x 2 x 1 = 120 ways.
5! = 120
The answer to this one is 24. You can do this mathematically by 4*3*2*1.
There are 2*5! = 240 ways.
The number of ways that 15 books can be arranged on a shelf is the same as the number of permutations of 15 things taken 15 at a time. This is 15 factorial, or 15!, and is 1,307,674,368,000.
it depends on the shelf of course
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
The answer depends on how many books on each subject there are.
The number of permutations of five things taken five at a time is five factorial, or 120.
5
5040
6! = 1 x 2 x 3 x 4 x 5 x 6.
7 factorial, or 5,040.
8 ! = 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1 = 40320
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.