## mathmath333 one year ago Counting problem

1. mathmath333

\large \color{black}{\begin{align}& \normalsize \text{The number of ways in which a Chairman and a Vice-Chairman can be chosen}\hspace{.33em}\\~\\ & \normalsize \text{from amongst a group of 12 persons assuming}\hspace{.33em}\\~\\ & \normalsize \text{that one person can not hold more than}\hspace{.33em}\\~\\ & \normalsize \text{one position.}\hspace{.33em}\\~\\ \end{align}}

2. ali2x2

this is off a google book xD

3. anonymous

do you know C notation?

4. ganeshie8

First choose chairman, there are $$12$$ ways of doing this. After that, choose vice-chairman, there are $$11$$ ways of doing this. So both can be done together in $$12\times 11$$ ways

5. anonymous

Using C notation:$$^{12}C_2\times2$$

6. mathmath333

what does it mean by \large \color{black}{\begin{align} & \normalsize \text{assuming that one person can not hold more}\hspace{.33em}\\~\\ & \normalsize \text{ than one position is ?}\hspace{.33em}\\~\\ \end{align}}

7. ganeshie8

Notice that the order in which you chose the ppl didnt matter here, end of the day all you care about is who were picked for chairman and vice-chairman, not the order in which you pick them

8. anonymous

It says one person can't be president and vice at the same time!

9. anonymous

Thank you @ganeshie8 and @mathmath333 for meddling me!!

10. ganeshie8

:) notice that $$^{12}C_2*2 = 12*11$$ which is same as the number of different strings of length $$2$$ formed by picking letters from a set of $$12$$ different letters. @mathmath333

11. mathmath333

also notice that $$\large ^{12}P_{2}=132$$

12. ganeshie8

Yep, all those interpretations are equivalent

13. ganeshie8

we have $\large ^nP_r~ ~=~~ ^nC_r * r!$