anonymous
  • 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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Without restrictions all the possibilities are accounted for by 18P5
anonymous
  • anonymous
but with the above restrictions.....:(
ganeshie8
  • ganeshie8
Wrong, does the order in which you pick people matter here ?

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anonymous
  • anonymous
I don't think so.
ganeshie8
  • 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
  • ganeshie8
So this better be a combination problem, yes ?
anonymous
  • anonymous
Yes.
ganeshie8
  • ganeshie8
Without restrictions all the possibilities are accounted for by 18\(\color{red}{C}\)5
anonymous
  • anonymous
What's the difference between P and C? I am sorry I am not very advanced
ganeshie8
  • ganeshie8
Maybe consider a quick example : Suppose you went to juice shop and ask for a salad made up of "bananas, apple, grapes"
ganeshie8
  • ganeshie8
does the order in which you ask matter here ?
anonymous
  • anonymous
No. You can go about picking any
anonymous
  • anonymous
But what difference does it make if permutation changed to combination?
anonymous
  • anonymous
Lack of order?
ganeshie8
  • 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
  • anonymous
ok I get it now.
ganeshie8
  • 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
  • ganeshie8
sorry to persist, but i think this is really important..
anonymous
  • anonymous
No
ganeshie8
  • ganeshie8
Are you saying "8 7 3 9" is no different from "7 3 8 9" ?
ganeshie8
  • ganeshie8
does your computer accept/treat both passwords as same ?
anonymous
  • anonymous
No
anonymous
  • anonymous
The computer only accepts a given password you made at the boot up
anonymous
  • anonymous
So the order would have to remain the same.
ganeshie8
  • ganeshie8
Exactly, since the "order" in which you enter the digits matter here, this is clearly a permutation problem.
ganeshie8
  • ganeshie8
That is how you differentiate between a permutation and combination problem
anonymous
  • anonymous
Oh ok. I get it now.
anonymous
  • anonymous
Ordinal or nominal
ganeshie8
  • 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
  • anonymous
24
ganeshie8
  • ganeshie8
good guess, but no.
anonymous
  • anonymous
Either this is too hard or I am too stupid:(
anonymous
  • anonymous
2.7 something
anonymous
  • anonymous
8/3=2.66
ganeshie8
  • 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
  • ganeshie8
where \(nCr\) is defined as \[nCr = \dfrac{n!}{r!*(n-r)!}\]
anonymous
  • anonymous
8C3=number of possibilities where 3 women are picked from 8 women.
ganeshie8
  • 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
  • ganeshie8
Yes, see if you can simplify above
anonymous
  • anonymous
(8*7*6*5)/3*2*1=336 1680/6=
anonymous
  • anonymous
Because most numbers cancel each other out.
anonymous
  • anonymous
8*7*6/3*2*1
ganeshie8
  • ganeshie8
Yes simplifies further
anonymous
  • anonymous
330/6=55
ganeshie8
  • ganeshie8
8*7*6/3*2*1 8*7 56 right ?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
so 56 ways in which 3 women can be picked from 8 women.
ganeshie8
  • ganeshie8
Yes next find how many ways 2 men can be picked from the available 10 men
anonymous
  • anonymous
2C10 so
ganeshie8
  • ganeshie8
10C2 ok
ganeshie8
  • ganeshie8
work it out
anonymous
  • anonymous
10!/2!(10-2)!
anonymous
  • anonymous
10!/2!*8!
anonymous
  • anonymous
10*9*8*7*6*5*4*3*2*1/2*1*8*7*6*5*4*3*2*1*
anonymous
  • anonymous
I simplify that
ganeshie8
  • ganeshie8
yes do it
anonymous
  • anonymous
90/2=45
anonymous
  • anonymous
45 ways in which 2 men can be picked from 10 men
anonymous
  • anonymous
56*45 is my guess.
ganeshie8
  • ganeshie8
Excellent! so a committee with 3 women and two men can be formed in : \[8C3*10C2 = 56*45~\text{ways}\]
anonymous
  • anonymous
I got it:)
anonymous
  • anonymous
Do I use the same for the last one?
anonymous
  • anonymous
where there are more men than women.
anonymous
  • 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
  • anonymous
Sorry reverse of that
anonymous
  • anonymous
more women than men. So do I multiply our previous answer by 3?
ganeshie8
  • ganeshie8
sry was away.. still here @Robert136
ganeshie8
  • 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
  • ganeshie8
you should get \[8C3*10C2 + 8C4*10C1+5C1*10C0\]
ganeshie8
  • 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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