anonymous
  • anonymous
a four person committee is chosen at random from a group of 15 people. how many different committees are possible?
Mathematics
chestercat
  • chestercat
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amistre64
  • amistre64
15 choose 4
amistre64
  • amistre64
15!/(4! 11!)
amistre64
  • amistre64
15.14.13.12.11! --------------- 4.3.2.11! 15.14.13.12 --------------- 4.3.2 15.7.13

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lgbasallote
  • lgbasallote
is this permutation? or combination @amistre64 ?
amistre64
  • amistre64
combo since it doesnt suggest positions for the 4
amistre64
  • amistre64
its just a group of 4 people with no specific need for position abc = bca in this case
lgbasallote
  • lgbasallote
i see thanks
anonymous
  • anonymous
where did you get the 15x7x13 ?
lgbasallote
  • lgbasallote
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
  • lgbasallote
do you get it @catgirl17 ?
anonymous
  • anonymous
where did the 6 come from in step 2?
ParthKohli
  • ParthKohli
\(\Large \color{MidnightBlue}{\Rightarrow nCr = {n! \over (n - r)!*r! } }\) Got it?
ParthKohli
  • ParthKohli
nCr can be represented as: \(\Large \color{MidnightBlue}{ (\matrix{n \\r}) }\)
anonymous
  • anonymous
no i don't get it @lgbasallote
lgbasallote
  • lgbasallote
hmm but you get 15.14.13.12 ----------- 4.3.2.1 right?
anonymous
  • anonymous
yes

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