Question 1157483: Question:a committee of 3 boys and 4 girls is to be formed from 5 boys and 6 girls. How many committees are possible if one girl refuses to serve if a certain other girl is chosen
My calculation :5c3 x 5c4 =50,but the answer given is 90
Answer by ikleyn(52864)   (Show Source):
You can put this solution on YOUR website! .
You should calculate the TOTAL possible number of combinations,
and then SUBTRACT from it the number of PROHIBITED combinations.
Does it work for you ?
Does it give you the right idea on how to move ahead ?
You may report me about your progress.
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On Combinations, see introductory lessons
- Introduction to Combinations
- PROOF of the formula on the number of Combinations
- Problems 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.
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Comment from student: Thanks for your reply but I can't get the idea to calculate the prohibited combination
My new calculation : (5c3 X 6c4)-(5c4x5c3)=100 Still didn't get the correct answer...
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My response :
The number of all combinations of 4G from 6G is  .
Of them, the prohibited combinations are those that contain these two anti-pathic girls.
The number of these prohibited combinations for girls is  :
you construct these prohibited combinations taking 2 antipathic girls,
and then you add to them 2 girls from remaining 6-2 = 4 "regular" girls.
So, for girls the number of prohibited teams is =  =  = 6.
You should combine them with all possible teams of boy, and it is what you should subtract from yours very first and most bigger number.
That's all.
Please let me know, if you do understand my explanations, and also whether you were able to complete the solution in full.
Happy solving (!)
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Comment from student : I have fully understand your explanation Thank you!!!!!
(5c3 x 6c4)-(5c3 x 4c2) = 90
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My response : You are absolutely right.
Your answer is correct !
My congratulations !
I am very glad to see your progress.
It is our common achievement: my in explanation, your in understanding.
To celebrate this breakthrough, I have a gift for you.
It is the list of additional (introductory) problems in elementary theory of probability in this site
- Simple and simplest probability problems
- Solving probability problems using complementary probability
- Elementary Probability problems related to combinations (*)
- A True/False test
- A multiple choice answers test
- Conditional probability problems
- Typical probability problems from the archive
- Experimental probability problems
- Elementary operations on sets help solving Probability problems
Although these problems are on Probability theory, they have many in common with Combinatorics problems.
In particular, the lesson, marked (*) in the list, contains many problems, that are close to the solved current problem.
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 "Solved problems on Probability".
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.
Consider these lessons as your textbook, handbook, a Solutions Manual, tutorials and (free of charge) home teacher.
Happy learning (!)
Come again to this forum soon to learn something new (!)
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