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)?

The answer is
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

Quote:

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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)]

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It is $\displaystyle 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 $\displaystyle 9\cdot 8$ ways. Then select either one French text or one German text to go with the two English books.
That can be done in $\displaystyle 9\cdot 8\cdot(7+5)$ ways. Then there are $\displaystyle 3$ places to put the single book.
That gives $\displaystyle 3\cdot[9\cdot 8\cdot(7+5)]$ 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.

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?

Hello, DBA!

Quote:

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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.)

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

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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: .$\displaystyle {9\choose2}{12\choose1} $ choices.
. . The three books can be arranged on the shelf in $\displaystyle 3!$ ways.

There are: .$\displaystyle 3!{9\choose2}{12\choose1}$ 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: .$\displaystyle {7\choose2}{14\choose1} $ choices.
. . The three books can be arranged on the shelf in $\displaystyle 3!$ ways.

There are: .$\displaystyle 3!{7\choose2}{14\choose1}$ 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: .$\displaystyle {5\choose2}{16\choose1} $ choices.
. . The three books can be arranged on the shelf in $\displaystyle 3!$ ways.

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


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

. . . . . . . . . . . $\displaystyle \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:

. . $\displaystyle 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.