SOLUTION: From a group of 6 girls and 7 boys, how many 5-member committees consist of : 3 girls and 2 boys? Multiple choice answers: a. 380 b. 240 c. 420 d. 400

Algebra ->  Permutations -> SOLUTION: From a group of 6 girls and 7 boys, how many 5-member committees consist of : 3 girls and 2 boys? Multiple choice answers: a. 380 b. 240 c. 420 d. 400       Log On


   



Question 1161688: From a group of 6 girls and 7 boys, how many 5-member committees consist of :
3 girls and 2 boys?

Multiple choice answers:
a. 380
b. 240
c. 420
d. 400

Found 2 solutions by KMST, ikleyn:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
There are 7%2A6 list of 2 boys that you could make from the 7 available boys,
but each group of 2 would appear in different order in
2 different lists,
so the number of sets of 2 boys that you can pick is
7%2A6%2F2=21 .

There are %286%2A5%2A4%29 list of 3 girls that you could make from the 6 available girls,
but each group of 3 girls would appear in different order in
3%2A2 different lists,
so the number of sets of 3 girls that you can pick is
%286%2A5%2A4%29%2F%283%2A2%29=20 .

Considering that the the choices of girls sets and boys sets are independent,
you would have 21%2A20=highlight%28420%29 different ways to form
5-member committees consisting of 3 girls and 2 boys.

Answer by ikleyn(52894) About Me  (Show Source):
You can put this solution on YOUR website!
.

You select 3 girls from 6 girls by  C%5B6%5D%5E3 = %286%2A5%2A4%29%2F%281%2A2%2A3%29 = 5*4 = 20 ways.


You select 2 boys from 7 boys by  C%5B7%5D%5E2 = %287%2A6%29%2F2 = 7*3 = 21 ways.


Since the selections are independent inside each group, the total number of ways is the product  20*21 = 420 ways.    ANSWER

Solved.

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See the lessons
    - Introduction to Combinations
    - PROOF of the formula on the number of Combinations
    - Problems on Combinations
    - OVERVIEW of lessons on Permutations and Combinations
    - Fundamental counting principle problems
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic  "Combinatorics: Combinations and permutations".


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Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.