## anonymous one year ago MEDAL!! (suppose there is a group of 7 people from which we will make a committee) In how many ways can we pick a three-person committee?

1. jim_thompson5910

Let's say we have 3 slots A,B,C how many do we have to choose from for slot A?

2. anonymous

7?

3. mathmate

Correct, then for slot B?

4. mathmate

Recall that one of the sever is already in slot A.

5. anonymous

6?

6. jim_thompson5910

yep, and then C?

7. anonymous

5

8. anonymous

multiply?

9. jim_thompson5910

yes, you'll multiply 7, 6 and 5

10. anonymous

210

11. jim_thompson5910

12. jim_thompson5910

210 is the number of ways to pick 3 people IF order mattered but there is no ranking on this committee, so order doesn't matter

13. jim_thompson5910

so what you have to do is divide by 3! = 3*2*1 = 6 to get the correct count

14. jim_thompson5910

the reason why is because there are 6 ways to order any three objects xyz xzy yxz yzx zxy zyx

15. anonymous

210 divided by 3?

16. jim_thompson5910

210/6 actually

17. jim_thompson5910

or 210/(3!)

18. anonymous

19. jim_thompson5910

correct

20. jim_thompson5910

another way is to use the combination formula $\Large _n C _r = \frac{n!}{r!*(n-r)!}$ with n = 7 and r = 3 and you'll get the same answer

21. anonymous

@jim_thompson5910 what if it was a four-person committee instead? wouldn't it be 7C4=35? also?

22. jim_thompson5910

that is correct

23. anonymous

thank you

24. jim_thompson5910

if you think of it in terms of 7 slots A through G we have this |dw:1441932178631:dw|

25. jim_thompson5910

now imagine cutting a line through that group such that one side (say the left side) has 3 slots and the other side has 4 slots |dw:1441932241059:dw|

26. jim_thompson5910

arranging 7 people to go in slots A through C is the exact same as arranging the remaining 4 to go in D through G so that's why 7 C 3 = 7 C 4 in general $\LARGE _n C _x = \ _n C _y$ where x+y = n (so in this case, x+y = 3+4 = 7 which is the value of n)

27. anonymous

Yes, that is the way my instructor explains it also, thanks again!

28. jim_thompson5910

you're welcome