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- anonymous

How many different 12-member juries can be chosen from a pool of 40 people?
I really can't understand this.

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- anonymous

- jamiebookeater

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- .Sam.

40C12?

- mathslover

And : \(\large ^n C_r = \cfrac{n!}{r! (n-r)!} \)

- mathslover

@Haruhi , can you solve it now?

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- anonymous

not really. I've been trying to solve it that way all this time. I'm getting weird results.

- .Sam.

From
\[\large ^n C_r = \cfrac{n!}{r! (n-r)!} = \frac{40!}{12!(40-12)!} = \frac{40 \times 39 \times 38 \times...}{(12 \times 11 \times 10 \times ...)(28 \times 27 \times 26 \times ...)}\]

- .Sam.

Or just use a calculator

- .Sam.

What do you get?

- anonymous

I get 1.5186643e+48 (whatever that means) and my options are
a) 3,586,853,480
b) 4,586,853,480
c) 5,586,853,480
d) 6,586,853,480

- .Sam.

Lol you should get 5,586,853,480 check your calculator

- Zarkon

type "calculate 40 choose 12" into google

- .Sam.

Ahh

- Zarkon

it will take
"40 choose 12"
and give the answer

- anonymous

I think we don't have Google in the exam!?

- anonymous

Ok..... still confused. How can a calculator be wrong lol what the hell am I doing

- anonymous

Try:\[\frac{40!}{(12!(40-12)!)} \]

- anonymous

That is wrong all of you I think its 12 times 40 its simple probability i think

- anonymous

No nevermind

- anonymous

Its c

- anonymous

It's the right answer, yeah. But it's damn confusing having to work with numbers like 815,915,279,999,999,744,392,488,672,368,848,648,400,776,424,296 -_- so I must've made a mistake at some point.

- anonymous

please dont say dam on a post there are 7th graders in geometry and algebra 2 like me who don't like that

- anonymous

ok... >_>

- anonymous

What is that supposed to mean

- anonymous

weirdos

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