- anonymous

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?

- chestercat

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- anonymous

Without restrictions all the possibilities are accounted for by 18P5

- anonymous

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

- ganeshie8

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

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## More answers

- anonymous

I don't think so.

- 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.

- ganeshie8

So this better be a combination problem, yes ?

- anonymous

Yes.

- ganeshie8

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

- anonymous

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

- ganeshie8

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

- ganeshie8

does the order in which you ask matter here ?

- anonymous

No. You can go about picking any

- anonymous

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

- anonymous

Lack of order?

- 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.

- anonymous

ok I get it now.

- 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 ?

- ganeshie8

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

- anonymous

No

- ganeshie8

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

- ganeshie8

does your computer accept/treat both passwords as same ?

- anonymous

No

- anonymous

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

- anonymous

So the order would have to remain the same.

- ganeshie8

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

- ganeshie8

That is how you differentiate between a permutation and combination problem

- anonymous

Oh ok. I get it now.

- anonymous

Ordinal or nominal

- 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 ?

- anonymous

24

- ganeshie8

good guess, but no.

- anonymous

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

- anonymous

2.7 something

- anonymous

8/3=2.66

- 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

- ganeshie8

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

- anonymous

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

- ganeshie8

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

- ganeshie8

Yes, see if you can simplify above

- anonymous

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

- anonymous

Because most numbers cancel each other out.

- anonymous

8*7*6/3*2*1

- ganeshie8

Yes simplifies further

- anonymous

330/6=55

- ganeshie8

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

- anonymous

yes.

- anonymous

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

- ganeshie8

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

- anonymous

2C10 so

- ganeshie8

10C2 ok

- ganeshie8

work it out

- anonymous

10!/2!(10-2)!

- anonymous

10!/2!*8!

- anonymous

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

- anonymous

I simplify that

- ganeshie8

yes do it

- anonymous

90/2=45

- anonymous

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

- anonymous

56*45 is my guess.

- ganeshie8

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

- anonymous

I got it:)

- anonymous

Do I use the same for the last one?

- anonymous

where there are more men than women.

- anonymous

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

- anonymous

Sorry reverse of that

- anonymous

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

- ganeshie8

sry was away.. still here @Robert136

- 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.

- ganeshie8

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

- ganeshie8

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

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