how many ways can u choose 3 people from a group of 6 boys n'5 girls IF there must be AT LEAST one boy chosen

Question

PERMUTATIONS, and (seperate question) IF there must be ONLY one girl chosen??

Answer 1

This is an exercise in the "Permutations" combination operator.

P(a,b) = a!/(a-b)!

There are P(11,3) = 11!/8! = 11*10*9 = 990 ways to choose a group of three people from a group of 11. (Note that this counts picking A-B-C as different from A-C-B even though the same group of three people results. However, the term "permutations" means that the ordering is significant and each different ordering is counted.)

There are P(5,3) = 5!/2! = 5*4*3 = 60 ways to choose three girls from a group of 5 girls.

There are therefore 990-60 = 930 ways to choose three people from six boys and five girls that "have at least one boy" (i.e., "do NOT have all girls"). I just took all possible permutations, and eliminated the all-girl ones.

You can't (as a subsequent responder did) assume that a boy MUST be chosen first.

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If exactly one girl is chosen, assuming we are still choosing three people from the same group, then you know you are choosing two boys and one girl. Then, there are:

C(6,2) = 6!/(2!*4!) = 6*5/2*1 = 15 different groups of two boys out of six boys

C(5,1) = 5!/(4!*1!) = 5/1 different "groups" of one girl out of five girls

Thus there are 15*5 = 75 different groups of two boys and one girl.

But there are six ways to pick each group (G-B1-B2, G-B2-B1, B1-G-B2, etc.), so there are 75*6 = 450 different ways to pick a group of three people that has exactly one girl, out of 6 boys and 5 girls.

Answer 2

1st person - must be a boy - 6 available choices
2nd person - can be anyone - thus 10 possible choices
3rd person - can be anyone - again 9 possible choice

thus 6 X 10 X 9 = 540.

Do the same thing for a girl

5 X 10 X 9 = 450

Answer 3

i'm assuming which you realize p - diversifications and c- mixture one million. C(10,5) - order would not remember on a committee 2. C(250,2) * C(one hundred,one million) - no order and multiply 3. C(13,6) - i'm assuming that order would not remember on a volleyball line up, yet i do no longer relatively understand. 4. C(3,one million) * C(11,4) 5. C(4,3) * C(6,4) 6. you may desire to apply the formulation, yet n = 10