Re: grouped probabilities

Hey lhurlbert.

If all the three books together then it means that you can look at each combination as "sliding" along for each possibility.

You have 8 books in total which means if you start all books at the left-most side you get 6 possibilities where the books start at positions 1,2,3,4,5 and 6.

Now we need to find the number of possibilities of re-arranging the books and for this we use a standard combinatoric identity.

The number of ways arranging 8 books given 5 engineering books (or 3 physics books) is 8C5 = 8C3 or using R:

> choose(8,5)

[1] 56

> choose(8,3)

[1] 56

So the total number of possibilities is 6/56 = 3/28 as expected.

Re: grouped probabilities

Hello, lhurlbert!

Quote:

Eight different book, 3 in Physics and 5 in Electrical Engineering, are placed at random on a shelf.

What is the probability that the three Physics books are all together?

Answer: 3/28

There are: .$\displaystyle 8! = 40,\!320$ possible orders.

Duct-tape the 3 Physics books together.

They can be ordered in $\displaystyle 3!=6$ ways.

We have 6 "books" to arrange: .$\displaystyle \boxed{ABC}\;D\;E\;F\;G\;H$

They can be ordered in $\displaystyle 6! = 720$ ways.

Hence, the 8 books can be ordered in $\displaystyle 6\cdot 720 = 4,\!320$ ways.

Therefore, the probability is: .$\displaystyle \frac{4,\!320}{40,\!320} \;=\;\frac{3}{28}$