Question

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

Answer 1

15 choose 4

Answer 2

15!/(4! 11!)

Answer 3

15.14.13.12.11!
---------------
4.3.2.11!


15.14.13.12
---------------
4.3.2


15.7.13

Answer 4

is this permutation? or combination @amistre64 ?

Answer 5

combo since it doesnt suggest positions for the 4

Answer 6

its just a group of 4 people with no specific need for position

abc = bca in this case

Answer 7

i see thanks

Answer 8

where did you get the 15x7x13 ?

Answer 9

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

Answer 10

do you get it @catgirl17 ?

Answer 11

where did the 6 come from in step 2?

Answer 12

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

Got it?

Answer 13

nCr can be represented as:


\(\Large \color{MidnightBlue}{ (\matrix{n \\r}) }\)

Answer 14

no i don't get it @lgbasallote

Answer 15

hmm but you get

15.14.13.12
-----------
4.3.2.1

right?

Answer 16

yes