nuber of combinations?

Question

A girl has four math books, three novels and two books of poetry. She wants to arrange them on her shelf so that all the math books are to the left, all the novels are in the middle, and all the poetry books are to the right. How many different arrangements of books are possible?

Answer 1

4P4*3P3*2P2
= 4!3!2!
= 24*6*2
= 288

that's the answer to "librarian arrangement".

if i arranged them "as long as their heights are vertical" with height > width > thickness, then i'll have to multiply the answer by itself 4 times over for the desired number of arrangement.

[4P4*3P3*2P2]^4
= 288^4
= 6879707136

Answer 2

Consider the number of combinations of each individual set.

How many ways can you arrange two books? Yep, two. How about three? Little more difficult, try to count it out. The answer is 6. Now, the most difficult one is 4, see if you can figure out the pattern (there is one! Look up the factorial on wikipedia).

For each of the single combinations of a set, there's the other ones that you can combine together, so if you have one set that has 3 combinations and another that has 2, then you can arrange the 2-set in one ways and then rearrange the 3-set 6 different ways. Then you can swap the 2-set and get an overall different combination, so 2*6 combinations in all.

Good luck!

Answer 3

The math books can be in any order and so can the novels and poetry books:

(4!) (3!) (2!) = 4x3x2x1x3x2x1x3x1= 432

Answer 4

One combination of groups.

24 combinations of math books
6 combinations of novels
2 combinations of poetry books

32 combinations of books
Taking into account the groups of books (math, novel, poetry) and the fact that those groups can only be arranged in one combination (math-left, novel-middle, poetry-right).

I WIN

Answer 5

Well... technically, if she has enough room on the shelf, she can arrange them with an infinite amount of combinations. But that would be stupid because it would involve moving all books or even just one book the smallest distance possible (perhaps a millimeter) then callinh the new position a new arrangement. Therefore, I suggest the following logical arrangements (all following the math-left, novels-middle, and poetry-right format):

All books stacked up with their binder edge facing outward with the lettering facing left.

All books stacked up with their binder edge facing outward with the lettering facing right.

All books stacked up with their binder edge facing inward with the lettering facing left.

All books stacked up with their binder edge facing inward with the lettering facing right.

All books stacked flat with their binder edge facing outward with the lettering facing up.

All books stacked flat with their binder edge facing outward with the lettering facing down.

All books stacked flat with their binder edge facing inward with the lettering facing up.

All books stacked flat with their binder edge facing inward with the lettering facing down.

Now you can imagine the number of ways you can change the orientation of each and every book yet still keep the format of math-left, novel-middle, and poetry-right! A lot.

Can't help you there.... good luck.

Peace

Answer 6

3! = 6

Answer 7

Think of it this way

LEFT MIDDLE RIGHT

---- * --- * --

ask how many choices you have for each place:

4*3*2*1 * 3*2*1 * 2*1

or 4!*3!*2!.

Do the rest. It's all multiplication.

Answer 8

There are 81 possibilities.

Answer 9

i would say about 24 different ways.

Answer 10

who knows