SOLUTION: A committee of six is to be chosen from the 20 members of the student council. In how many ways can this be done if: A. There are no restrictions. B. There are twelve junior

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Question 1189221: A committee of six is to be chosen from the 20 members of the student council. In how many ways can this be done if:
A. There are no restrictions.
B. There are twelve junior members and 8 senior members and the committee must include 2 junior and four senior members?
C. There must be at least 2 senior members on the committee. ( I need help with this one in particular)

Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
A committee of six is to be chosen from the 20 members of the student council.
In how many ways can this be done if:
A. There are no restrictions.
B. There are twelve junior members and 8 senior members and the committee must include 2 junior and four senior members?
C. There must be at least 2 senior members on the committee. ( I need help with this one in particular)
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            As you request,  I solve the problem's  Part  C   only. 


"at least 2 senior members" means 2, or 3, or 4, or 5, or 6 senior members.


So, the formula is


    in how many ways =  = 

                     = 28*495 + 56*220 + 70*66 + 56*12 + 28*1 = 31500 ways.    ANSWER

Solved.

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Post-solution notes


        (1)   The formula is self-explanatory . . .

        (2)   It is tedious to calculate the binomial coefficients  (combinations)  manually - - -
                - - - so I used the  Excel standard function  combin(n,k)  as a calculator.


To make your horizon wider and the ground under your legs solder,  see the lessons
    - Introduction to Combinations
    - PROOF of the formula on the number of Combinations
    - Problems on Combinations
    - Problems on Combinations with restrictions
    - Fundamental counting principle problems
    - Some twisted combinatorics problem
    - Math circle level problem on Combinations
    - OVERVIEW of lessons on Permutations and Combinations
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".


Save the link to this textbook together with its description

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.



Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Part A

Each seat on a committee is equal in rank to any other seat, since none of the seats are labeled (eg: chairman, president, VP, etc). This tells us that order doesn't matter. A committee like {Alice,Bob,Christy} is the same as the committee {Bob,Christy,Alice}. All that matters is the overall group rather than the individual members themselves.

Since order doesn't matter, we use the nCr combination formula
C%28n%2Cr%29+=+%28n%21%29%2F%28r%21%2A%28n-r%29%21%29
The exclamation marks indicate factorials.

We plug in n = 20 and r = 6

C%28n%2Cr%29+=+%28n%21%29%2F%28r%21%2A%28n-r%29%21%29

C%2820%2C6%29+=+%2820%21%29%2F%286%21%2A%2820-6%29%21%29

C%2820%2C6%29+=+%2820%21%29%2F%286%21%2A14%21%29

C%2820%2C6%29+=+%2820%2A19%2A18%2A17%2A16%2A15%2A14%21%29%2F%286%21%2A14%21%29



The "14!" terms cancel

C%2820%2C6%29+=+%2820%2A19%2A18%2A17%2A16%2A15%29%2F%286%21%29

C%2820%2C6%29+=+%2820%2A19%2A18%2A17%2A16%2A15%29%2F%286%2A5%2A4%2A3%2A2%2A1%29

C%2820%2C6%29+=+%28%2227%2C907%2C200%22%29%2F%28720%29

C%2820%2C6%29+=+%2238%2C760%22

There are 38,760 ways to form the six person committee from a pool of 20 students. Order doesn't matter.

Answer: 38,760

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Part B

We'll use the nCr formula again to find out how many ways there are to pick the juniors (ignore the seniors for now).
There is a pool of n = 12 juniors and we want to select r = 2 of them.
Using the nCr formula, you should find that C(n,r) = C(12,2) = 66. I'll skip the steps in computing the nCr value from now on since the steps are similar to part A. Let me know if you need to see the steps.

Anyways, because C(12,2) = 66, this means there are 66 ways to pick the juniors.

Also, there are C(8,4) = 70 ways to pick the seniors. Use n = 8 and r = 4.

Overall, there are 66*70 = 4620 ways to pick two juniors and four seniors from a pool of 12 and 8 respectively.

Answer: 4620

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Part C

Consider these cases:
Case A) Picking 6 juniors + 0 seniors
Case B) Picking 5 juniors + 1 senior

For case A, there are C(12,6) = 924 ways to pick six juniors and C(8,0) = 1 way to pick zero seniors. So there are A = 924*1 = 924 ways to satisfy case A.

For case B, we have C(12,5) = 792 ways to pick five juniors and C(8,1) = 8 ways to pick one senior. This leads to B = 792*8 = 6336 ways to select the committe to satisfy case B.

Adding the results so far
A+B = 924+6336 = 7260
This represents the number of ways to form a committee where we have less than 2 seniors.
The complement of "less than 2 seniors" is "at least 2 seniors" (aka 2 or more).

Back in part A, we computed the result 38,760 that counted the number of ways to form a six-person committee regardless of restriction (ie quota counts for each grade level). We'll subtract the 7260 figure from 38,760 to get the final answer

38,760 - 7260 = 31,500

Answer: 31,500