# Homework Help: Probability arranging books

1. Nov 19, 2012

### mtingt

1. The problem statement, all variables and given/known data
there are 7 different french books and 7 different Spanish books, how many ways are there to arrange them on a shelf
a. books of the same language must be group together, French on left and Spanish on Right?
b. French and Spanish books must alternate in the grouping, beginning with a French book?

I tried doing 7!x7! for both of them but i don't think i am right?

I have no idea how to approach this

2. Nov 19, 2012

### haruspex

Sounds right to me.

3. Nov 22, 2012

for the first one 7!x7! seems right, but for the 2nd one I think (not sure...!!!) it's :

7C1X7C1 X 6C1X6C1 X 5C1X5C1 X 4C1X4C1 X 3C1X3C1 X 2C1X2C1 X 1C1X1C1

4. Nov 22, 2012

### CAF123

I think an argument could go: there are 14 choices for the first book, (French or Spanish). There are then 7 choices for the next book (If first was French, this one must be Spanish), then 6 choices for next, (has to be French), then 6 choices for next (has to be Spanish)....and so on. In this Q, the French book is first so what you wrote is correct.

Last edited: Nov 22, 2012
5. Nov 22, 2012

### Ray Vickson

Another argument: for each arrangement of the French books, leave a space between successive books and fill those spaces with the Spanish books, one book per space.

RGV

Last edited: Nov 22, 2012
6. Nov 22, 2012

### haruspex

How is that different from 7!x7!?
The two obviously have the same answer. Either way, there is a fixed set of 7 positions that can be taken by the French books, and another fixed set of 7 that can be taken by the Spanish, independently.

7. Nov 22, 2012

### Ray Vickson

It's not different; it's just another argument that the OP may, or may not, prefer.

RGV

8. Nov 22, 2012