Basic Counting Principles

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September 23rd 2010, 07:10 AM
DBA

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
September 23rd 2010, 07:52 AM
Plato

Quote:

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.
September 23rd 2010, 07:54 AM
aman_cc

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?
September 23rd 2010, 09:35 AM
Soroban

Hello, DBA!

Quote:

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.