## anonymous one year ago HELP PLEASE:) FAN&MEDAL&TESTIMONIAL:):) A committee of 5 people is to be chosen from a group of 8 women and 10 men. The committee must feature 3 women and two men? The committee must have more women than men?

1. anonymous

Without restrictions all the possibilities are accounted for by 18P5

2. anonymous

but with the above restrictions.....:(

3. ganeshie8

Wrong, does the order in which you pick people matter here ?

4. anonymous

I don't think so.

5. ganeshie8

The order in which you select the people doesn't matter here because, end of the day, all you bother about is "who" are in the committee. Not in which order you picked them.

6. ganeshie8

So this better be a combination problem, yes ?

7. anonymous

Yes.

8. ganeshie8

Without restrictions all the possibilities are accounted for by 18$$\color{red}{C}$$5

9. anonymous

What's the difference between P and C? I am sorry I am not very advanced

10. ganeshie8

Maybe consider a quick example : Suppose you went to juice shop and ask for a salad made up of "bananas, apple, grapes"

11. ganeshie8

does the order in which you ask matter here ?

12. anonymous

No. You can go about picking any

13. anonymous

But what difference does it make if permutation changed to combination?

14. anonymous

Lack of order?

15. ganeshie8

give me a salad made up of "bananas, apple, grapes" is no different from give me a salad made up of "apple, bananas, grapes" order doesn't matter here, so this is a combination problem.

16. anonymous

ok I get it now.

17. ganeshie8

consider another quick example : you want to choose four digits to set a password to your laptop hard disk : "8 7 3 9" does the order matter here ?

18. ganeshie8

sorry to persist, but i think this is really important..

19. anonymous

No

20. ganeshie8

Are you saying "8 7 3 9" is no different from "7 3 8 9" ?

21. ganeshie8

22. anonymous

No

23. anonymous

The computer only accepts a given password you made at the boot up

24. anonymous

So the order would have to remain the same.

25. ganeshie8

Exactly, since the "order" in which you enter the digits matter here, this is clearly a permutation problem.

26. ganeshie8

That is how you differentiate between a permutation and combination problem

27. anonymous

Oh ok. I get it now.

28. anonymous

Ordinal or nominal

29. ganeshie8

 A committee of 5 people is to be chosen from a group of 8 women and 10 men. The committee must feature 3 women and two men?  How many ways can you choose 3 women from 8 women ?

30. anonymous

24

31. ganeshie8

good guess, but no.

32. anonymous

Either this is too hard or I am too stupid:(

33. anonymous

2.7 something

34. anonymous

8/3=2.66

35. ganeshie8

Neither, you just don't happen to have enough practice. You can think of this exactly same as the permutation problem. 3 women can be chosen from 8 women in 8C3 ways

36. ganeshie8

where $$nCr$$ is defined as $nCr = \dfrac{n!}{r!*(n-r)!}$

37. anonymous

8C3=number of possibilities where 3 women are picked from 8 women.

38. ganeshie8

$8C3 = \dfrac{8!}{3!*(8-3)!} = \dfrac{8!}{3!*5!}=\dfrac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1*5\times 4\times 3\times 2\times 1}$

39. ganeshie8

Yes, see if you can simplify above

40. anonymous

(8*7*6*5)/3*2*1=336 1680/6=

41. anonymous

Because most numbers cancel each other out.

42. anonymous

8*7*6/3*2*1

43. ganeshie8

Yes simplifies further

44. anonymous

330/6=55

45. ganeshie8

8*7*6/3*2*1 8*7 56 right ?

46. anonymous

yes.

47. anonymous

so 56 ways in which 3 women can be picked from 8 women.

48. ganeshie8

Yes next find how many ways 2 men can be picked from the available 10 men

49. anonymous

2C10 so

50. ganeshie8

10C2 ok

51. ganeshie8

work it out

52. anonymous

10!/2!(10-2)!

53. anonymous

10!/2!*8!

54. anonymous

10*9*8*7*6*5*4*3*2*1/2*1*8*7*6*5*4*3*2*1*

55. anonymous

I simplify that

56. ganeshie8

yes do it

57. anonymous

90/2=45

58. anonymous

45 ways in which 2 men can be picked from 10 men

59. anonymous

56*45 is my guess.

60. ganeshie8

Excellent! so a committee with 3 women and two men can be formed in : $8C3*10C2 = 56*45~\text{ways}$

61. anonymous

I got it:)

62. anonymous

Do I use the same for the last one?

63. anonymous

where there are more men than women.

64. anonymous

there are 3 cases where men can be more than women 3 men 2 women 4 men 1 woman 5 men 0 women

65. anonymous

Sorry reverse of that

66. anonymous

more women than men. So do I multiply our previous answer by 3?

67. ganeshie8

sry was away.. still here @Robert136

68. ganeshie8

 there are 3 cases where men can be more than women 3 women 2 men 4 women 1 man 5 women 0 man  Yes find the number of ways for each case, then add up.

69. ganeshie8

you should get $8C3*10C2 + 8C4*10C1+5C1*10C0$

70. ganeshie8