permutation

• Sep 12th 2008, 07:59 AM
permutation
there were 15 english books ,9 spanish books,17 french books on a shelf.How many ways can you pick two books without them being the same.
• Sep 12th 2008, 09:01 AM
Soroban

Quote:

There are 15 English books, 9 Spanish books, 17 French books on a shelf.
In how many ways can you pick two books without them being the same?

$\displaystyle \begin{array}{cccc}\text{Ways to pick one English and one Spanish:} & 15\cdot9 &=& 135 \\ \text{Ways to pick one Spanish and one French:} & 9 \cdot 17 &=& 153 \\ \text{Ways to pick one English and one French:} & 15\cdot17 &=& 255 \end{array}$

Answer: .$\displaystyle 135 + 153 + 255 \;=\;\boxed{543}$ ways to get two different languages.

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Check

There are: .$\displaystyle {41\choose2} \:=\:820$ possible choices of two books.

$\displaystyle \begin{array}{ccccc}\text{There are:} &{15\choose2} &=& 105 &\text{ways to get 2 English books} \\ \\[-3mm] \text{There are:}& {9\choose2} &=& 36 & \text{ways to get 2 Spanish books} \\ \\[-3mm] \text{There are:} & {17\choose2} &=& 136 & \text{ways to get 2 French books} \end{array}$

Hence, there are: .$\displaystyle 105 + 36 + 136 \:=\:277$ ways to get the same language.

Therefore, there are: .$\displaystyle 820 - 277 \:=\:\boxed{543}$ ways to get different languages.

By the way, these are not permuations.
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