catgirl17
a four person committee is chosen at random from a group of 15 people. how many different committees are possible?



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amistre64
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15 choose 4

amistre64
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15!/(4! 11!)

amistre64
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15.14.13.12.11!

4.3.2.11!
15.14.13.12

4.3.2
15.7.13

lgbasallote
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is this permutation? or combination @amistre64 ?

amistre64
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combo since it doesnt suggest positions for the 4

amistre64
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its just a group of 4 people with no specific need for position
abc = bca in this case

lgbasallote
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i see thanks

catgirl17
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where did you get the 15x7x13 ?

lgbasallote
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15.14.13.12

4.3.2.1
15.14.13.6

4.3.
15.14.13.2

4
15.14.13

2
15.7.13

lgbasallote
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do you get it @catgirl17 ?

catgirl17
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where did the 6 come from in step 2?

ParthKohli
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\(\Large \color{MidnightBlue}{\Rightarrow nCr = {n! \over (n  r)!*r! } }\)
Got it?

ParthKohli
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nCr can be represented as:
\(\Large \color{MidnightBlue}{ (\matrix{n \\r}) }\)

catgirl17
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no i don't get it @lgbasallote

lgbasallote
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hmm but you get
15.14.13.12

4.3.2.1
right?

catgirl17
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yes