A community for students.
Here's the question you clicked on:
 0 viewing
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?
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?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Without restrictions all the possibilities are accounted for by 18P5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but with the above restrictions.....:(

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Wrong, does the order in which you pick people matter here ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4The 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
 one year ago
Best ResponseYou've already chosen the best response.4So this better be a combination problem, yes ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Without restrictions all the possibilities are accounted for by 18\(\color{red}{C}\)5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's the difference between P and C? I am sorry I am not very advanced

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Maybe consider a quick example : Suppose you went to juice shop and ask for a salad made up of "bananas, apple, grapes"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4does the order in which you ask matter here ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No. You can go about picking any

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But what difference does it make if permutation changed to combination?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4`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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4consider 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
 one year ago
Best ResponseYou've already chosen the best response.4sorry to persist, but i think this is really important..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Are you saying "8 7 3 9" is no different from "7 3 8 9" ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4does your computer accept/treat both passwords as same ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The computer only accepts a given password you made at the boot up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the order would have to remain the same.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Exactly, since the "order" in which you enter the digits matter here, this is clearly a permutation problem.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4That is how you differentiate between a permutation and combination problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok. I get it now.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4``` 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
 one year ago
Best ResponseYou've already chosen the best response.0Either this is too hard or I am too stupid:(

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Neither, 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
 one year ago
Best ResponseYou've already chosen the best response.4where \(nCr\) is defined as \[nCr = \dfrac{n!}{r!*(nr)!}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.08C3=number of possibilities where 3 women are picked from 8 women.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[8C3 = \dfrac{8!}{3!*(83)!} = \dfrac{8!}{3!*5!}=\dfrac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1*5\times 4\times 3\times 2\times 1}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes, see if you can simplify above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(8*7*6*5)/3*2*1=336 1680/6=

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because most numbers cancel each other out.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes simplifies further

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.48*7*6/3*2*1 8*7 56 right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 56 ways in which 3 women can be picked from 8 women.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes next find how many ways 2 men can be picked from the available 10 men

anonymous
 one year ago
Best ResponseYou've already chosen the best response.010*9*8*7*6*5*4*3*2*1/2*1*8*7*6*5*4*3*2*1*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.045 ways in which 2 men can be picked from 10 men

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Excellent! so a committee with 3 women and two men can be formed in : \[8C3*10C2 = 56*45~\text{ways}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do I use the same for the last one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where there are more men than women.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there are 3 cases where men can be more than women 3 men 2 women 4 men 1 woman 5 men 0 women

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry reverse of that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0more women than men. So do I multiply our previous answer by 3?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4sry was away.. still here @Robert136

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4``` 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
 one year ago
Best ResponseYou've already chosen the best response.4you should get \[8C3*10C2 + 8C4*10C1+5C1*10C0\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4wolfram 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.