Factorials/Permutations

wquorh

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Feb 12, 2006
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How many ways can you pick a president, vice president and secretary from a group of six boys and five girls if there must be only one girl chosen?

First I calculated 11P5 = 990, and now I'm kind of stuck. How should I implement the restriction stated in the qeustion? If I subtract 6P3 (120) from the answer, I'm assuming that this will give me the answer to so that there are at LEAST one girl chosen right? How can I find the answer if "only one girl" is chosen? Thanks a lot. :D
 
First, let \(\displaystyle \L
\left( \begin{array}{l}
N \\
k \\
\end{array} \right) = \frac{{N!}}{{k!\left( {N - k} \right)!}}\) be the combination of N choosing k.

The number choices is \(\displaystyle \L
\left( \begin{array}{l}
6 \\
2 \\
\end{array} \right)\left( \begin{array}{l}
5 \\
1 \\
\end{array} \right)\).

Now we order them: \(\displaystyle \L
\left( \begin{array}{l}
6 \\
2 \\
\end{array} \right)\left( \begin{array}{l}
5 \\
1 \\
\end{array} \right)\left( {3!} \right)\).
 
Ok, so I'm still sorta confused. I thought this was permutations not combinations? Would this make sense? \(\displaystyle 6P2 \times 5P1\), so..

\(\displaystyle (\frac{6!}{4!}) \times (\frac{5!}{4!})\)??

Why, in your solution, do you multiply by 3!? Sorry if what I'm saying doesn't make much sense.. I just started the semester and we are learning about permutaions.[/tex]
 
Well first we choose two boys and one girl, say {Adam, James, Trisha}
There are 75 different such sets
Then for each such set there are (3!) ways to assign Pres., VP, and Sec.
Example: {Adam-Sec, James-Pres, Trisha-VP} or maybe {Adam-VP, James-sec, Trisha-Pres}, etc.
 
Ahhh I get it now. I guess because we didn't learn about combinations yet I got confused. But yeah, makes a lot of sense, thanks :D
 
Hello, wquorh!

How many ways can you pick a president, vice president and secretary
from a group of six boys and five girls if there must be only one girl chosen?
First, select the one girl . . . there are 5 choices.
Then give her a position . . . there are 3 choices.

The other two positions are filled by boys: \(\displaystyle 6\,\times\,5\,=\,30\) choices.

Therefore, there are: \(\displaystyle \,5\,\times\,3\,\times\,30\:=\:450\) ways.
 
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