1. ## Basic Counting Principles

Given nine different English books, seven different French books, and five different German books.

How many ways are there to make a row of three books in which exactly one language is missing (the order of the three books makes a difference)?

3*[9*8* (7+5) + 7*6*(9+5) + 5*4*(9+7)]

I do not understand how to get there.

I have 3 locations in the row _ _ _

In the case I use only French and German books I have
7+5=12 books availabe to place
I do not kow where the 9*8 come from in the expression 3*(9*8*(7+5))

I seem to have 3 subsets which are added (French/German, Engl/German, Engl/French).

Can someone help me out? Thanks

2. Originally Posted by DBA
Given nine different English books, seven different French books, and five different German books.
How many ways are there to make a row of three books in which exactly one language is missing (the order of the three books makes a difference)?
The answer is 3*[9*8* (7+5) + 7*6*(9+5) + 5*4*(9+7)]
It is $3\cdot[9\cdot8\cdot (7+5) + 7\cdot6\cdot(9+5) + 5\cdot4\cdot(9+7)]$

Consider the case in which we use two English texts. That can happen in $9\cdot 8$ ways. Then select either one French text or one German text to go with the two English books.
That can be done in $9\cdot 8\cdot(7+5)$ ways. Then there are $3$ places to put the single book.
That gives $3\cdot[9\cdot 8\cdot(7+5)]$ to form a line of two English texts, one French text, or a line of two English texts, one German text.
Now you have two more cases.

3. Some hint -
"In the case I use only French and German books I have
7+5=12 books availabe to place"

But there is an added condition - there has to be atleast 1 book of French and German each.

Can you work fwd now?

4. Hello, DBA!

Given: 9 distinct English books, 7 distinct French books, and 5 distinct German books.

How many ways are there to make a row of three books
in which exactly one language is missing?
(The order of the three books makes a difference.)

$\text{The answer is: }\;3\cdot\bigg[9\cdot8\cdot (7+5) + 7\cdot6\cdot(9+5) + 5\cdot4\cdot(9+7)\bigg]$

Their answer is correct . . . but why did they write it that way??

Suppose we want two English books and one Other.
. . There are 9 English and 12 Others.
. . There are: . ${9\choose2}{12\choose1}$ choices.
. . The three books can be arranged on the shelf in $3!$ ways.

There are: . $3!{9\choose2}{12\choose1}$ ways to have two English books and one Other.

Suppose we want two French books and one Other.
. . There are 7 French and 14 Others.
. . There are: . ${7\choose2}{14\choose1}$ choices.
. . The three books can be arranged on the shelf in $3!$ ways.

There are: . $3!{7\choose2}{14\choose1}$ ways to have two French books and one Other.

Suppose we want two German books and one Other.
. . There are 5 German and 16 Others.
. . There are: . ${5\choose2}{16\choose1}$ choices.
. . The three books can be arranged on the shelf in $3!$ ways.

There are: . $3!{5\choose2}{16\choose1}$ ways to have two German books and one Other.

Hence, there are: . $\displaystyle 3!{9\choose2}{12\choose1} + 3!{7\choose2}{14\choose1} + 3!{5\choose2}{16\choose1}$

. . . . . . . . . . . $\displaystyle =\;3!\bigg[{9\choose2}\!\cdot\112 + {7\choose2}\!\cdot\!14 + {5\choose2}\!\cdot\116\bigg]\text{ ways}$ .[1]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

At this point, I would have cranked out the answer:

. . $6\cdot\bigg[\frac{9\cdot8}{2}\cdot12 + \frac{7\cdot6}{2}\cdot14 + \frac{5\cdot4}{2}\cdot16\bigg] \;=\;6(886) \;=\;5316$

Or left it as [1] so others might see how I got the answer.

Their "3" out front can be explained,
. . but I think it is confusing and misleading.