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

Question

amistre64 · Answer

15 choose 4

amistre64 · Answer

15!/(4! 11!)

amistre64 · Answer

15.14.13.12.11!
---------------
     4.3.2.11!


15.14.13.12
---------------
      4.3.2


15.7.13

lgbasallote · Answer

is this permutation? or combination @amistre64 ?

amistre64 · Answer

combo since it doesnt suggest positions for the 4

amistre64 · Answer

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

lgbasallote · Answer

i see thanks

anonymous · Answer

where did you get the 15x7x13 ?

lgbasallote · Answer

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 · Answer

do you get it @catgirl17 ?

anonymous · Answer

where did the 6 come from in step 2?

parthkohli · Answer

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

parthkohli · Answer

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

anonymous · Answer

no i don't get it @lgbasallote

lgbasallote · Answer

hmm but you get 

15.14.13.12
-----------
4.3.2.1

right?

anonymous · Answer

yes