Question

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?

Answer 1

Without restrictions all the possibilities are accounted for by 18P5

Answer 2

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

Answer 3

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

Answer 4

I don't think so.

Answer 5

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.

Answer 6

So this better be a combination problem, yes ?

Answer 7

Yes.

Answer 8

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

Answer 9

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

Answer 10

Maybe consider a quick example :

Suppose you went to juice shop and ask for a salad made up of "bananas, apple, grapes"

Answer 11

does the order in which you ask matter here ?

Answer 12

No. You can go about picking any

Answer 13

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

Answer 14

Lack of order?

Answer 15

`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.

Answer 16

ok I get it now.

Answer 17

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 ?

Answer 18

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

Answer 19

No

Answer 20

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

Answer 21

does your computer accept/treat both passwords as same ?

Answer 22

No

Answer 23

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

Answer 24

So the order would have to remain the same.

Answer 25

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

Answer 26

That is how you differentiate between a permutation and combination problem

Answer 27

Oh ok. I get it now.

Answer 28

Ordinal or nominal

Answer 29

```
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 ?

Answer 30

24

Answer 31

good guess, but no.

Answer 32

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

Answer 33

2.7 something

Answer 34

8/3=2.66

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

Yes, see if you can simplify above

Answer 40

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

Answer 41

Because most numbers cancel each other out.

Answer 42

8*7*6/3*2*1

Answer 43

Yes simplifies further

Answer 44

330/6=55

Answer 45

8*7*6/3*2*1

8*7

56

right ?

Answer 46

yes.

Answer 47

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

Answer 48

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

Answer 49

2C10 so

Answer 50

10C2 ok

Answer 51

work it out

Answer 52

10!/2!(10-2)!

Answer 53

10!/2!*8!

Answer 54

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

Answer 55

I simplify that

Answer 56

yes do it

Answer 57

90/2=45

Answer 58

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

Answer 59

56*45 is my guess.

Answer 60

Excellent! so a committee with 3 women and two men can be formed in :

\[8C3*10C2 = 56*45~\text{ways}\]

Answer 61

I got it:)

Answer 62

Do I use the same for the last one?

Answer 63

where there are more men than women.

Answer 64

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

Answer 65

Sorry reverse of that

Answer 66

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

Answer 67

sry was away.. still here @Robert136

Answer 68

```
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.

Answer 69

you should get
\[8C3*10C2 + 8C4*10C1+5C1*10C0\]

Answer 70

wolfram says that evaluates to 3225

http://www.wolframalpha.com/input/?i=%288+choose+3%29*%2810+choose+2%29+%2B+%288+choose+4%29*%2810+choose+1%29%2B%285+choose+1%29*%2810+choose+0%29