How Many Ways to Form Groups from Boys and Girls in a Classroom?

AI Thread Summary
To solve the problem of forming groups from 15 boys and 19 girls, the first part requires calculating the number of groups with exactly four girls and five boys, which is done using combinations: C(19,4) * C(15,5) = 11639628. For part b, the goal is to form a group of 14 with an equal number of boys and girls, which means there would be 7 boys and 7 girls. Part c involves determining the number of groups of 5 that have more boys than girls, which can be achieved by considering combinations for scenarios where the number of boys exceeds the number of girls, such as 5-0, 4-1, and 3-2. Overall, the discussion emphasizes the application of combinatorial mathematics to group formation problems.
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Homework Statement


There are 15 boys and 19 girls in a room.
a) Find the number of different groups that contain exactly four girls and five boys.
b) How many groups of 14 have an equal number of boys and girls?
c) How many groups of 5 have more boys than girls?


Homework Equations





The Attempt at a Solution


a) C(19,4) * C(15,5)
=11639628

b & c I have no idea where to start.
 
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Isn't b just like a? If you have 14 people with equal numbers of boys and girls then how many boys? Girls?

And for c, what are the choices that qualify from 5-0, 4-1, 3-2, 2-3, 1-4, 0-5. Figure them out and add them up.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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