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

Mathematics
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40C12?
And : \(\large ^n C_r = \cfrac{n!}{r! (n-r)!} \)
@Haruhi , can you solve it now?

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Other answers:

not really. I've been trying to solve it that way all this time. I'm getting weird results.
From \[\large ^n C_r = \cfrac{n!}{r! (n-r)!} = \frac{40!}{12!(40-12)!} = \frac{40 \times 39 \times 38 \times...}{(12 \times 11 \times 10 \times ...)(28 \times 27 \times 26 \times ...)}\]
Or just use a calculator
What do you get?
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
Lol you should get 5,586,853,480 check your calculator
type "calculate 40 choose 12" into google
Ahh
it will take "40 choose 12" and give the answer
I think we don't have Google in the exam!?
Ok..... still confused. How can a calculator be wrong lol what the hell am I doing
Try:\[\frac{40!}{(12!(40-12)!)} \]
That is wrong all of you I think its 12 times 40 its simple probability i think
No nevermind
Its c
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.
please dont say dam on a post there are 7th graders in geometry and algebra 2 like me who don't like that
ok... >_>
What is that supposed to mean
weirdos

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