Question

Counting problem

Answer 1

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

Answer 2

this is off a google book xD

Answer 3

do you know C notation?

Answer 4

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

Answer 5

Using C notation:\(^{12}C_2\times2\)

Answer 6

what does it mean by

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

:) 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

Answer 11

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

Answer 12

Yep, all those interpretations are equivalent

Answer 13

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