## catgirl17 3 years ago a four person committee is chosen at random from a group of 15 people. how many different committees are possible?

1. amistre64

15 choose 4

2. amistre64

15!/(4! 11!)

3. amistre64

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

4. lgbasallote

is this permutation? or combination @amistre64 ?

5. amistre64

combo since it doesnt suggest positions for the 4

6. amistre64

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

7. lgbasallote

i see thanks

8. catgirl17

where did you get the 15x7x13 ?

9. 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

10. lgbasallote

do you get it @catgirl17 ?

11. catgirl17

where did the 6 come from in step 2?

12. ParthKohli

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

13. ParthKohli

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

14. catgirl17

no i don't get it @lgbasallote

15. lgbasallote

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

16. catgirl17

yes