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permutation

by jzw » Fri Feb 24, 2012 6:11 pm
There is a group of 4 boys and 6 girls. How many study groups of 3 can be formed such that one member of the group is a boy? (two groups are considered different if @ least one member of the group is different.)

Drawing it out takes forever but I got 100.

How do I set this one up as a formula?
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by sanju09 » Sat Feb 25, 2012 1:11 am
jzw wrote:There is a group of 4 boys and 6 girls. How many study groups of 3 can be formed such that one member of the group is a boy? (two groups are considered different if @ least one member of the group is different.)

Drawing it out takes forever but I got 100.

How do I set this one up as a formula?
Please revise Permutation, Combination, and few more counting techniques before writing test.

This question unequivocally wants us to select 1 boy from 4 and 2 girls from 6 to form a study group of 3.

This can be done in 4C1 × 6C2 = [spoiler]4 × 15 = 60[/spoiler].

I am not getting 100.
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by GMATGuruNY » Sat Feb 25, 2012 3:41 am
jzw wrote:There is a group of 4 boys and 6 girls. How many study groups of 3 can be formed such that one member of the group is a boy? (two groups are considered different if @ least one member of the group is different.)

Drawing it out takes forever but I got 100.

How do I set this one up as a formula?
I received a PM asking me to comment.

This is what I call a "bucket" problem.
We have two buckets: a bucket of boys and bucket of girls.
We have to make choices from each bucket.

Step 1: Count the number of options from each bucket.
The number of ways to choose one boy from 4 choices = 4.
The number of ways to choose a combination of 2 girls from 6 choices = (6*5)/(2*1) = 15.

Step 2: To combine the number of options from each bucket, multiply.
4*15 = 60.
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by ubhanja » Sat Feb 25, 2012 7:25 am
Is it at least 1 boy or maximum 1 boy.

If it is maximum one boy the answer provided by Mitch is right.

If it is at least one boy; I feel it should be 4C1 * 9C2 [6 girls and 1 boy] =144.

Please clarify.
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by tomada » Sat Feb 25, 2012 11:08 am
First I started with the total number of ways to choose a study group of 3 people from 10 people: 10C3 = 10!/(7!)(3!) = 120.

This number, 120, can be thought of as the sum of the following:

# of ways a 3-person group can be formed containing zero boys +
# of ways a 3-person group can be formed containing one boy +
# of ways a 3-person group can be formed containing two boys +
# of ways a 3-person group can be formed containing three boys

The above comprise an exhaustive set of possibilities which, as shown by 10C3, is 120.

The phrase at least one boy is represented by the lattermost three of the four possible scenarios shown above.

So, if we can calculate the first scenario - namely, the # of ways a 3-person group can be formed containing zero boys,
we can subtract that number from 120, and we'll have our answer.

How many ways are there to form a 3-person group consisting solely of girls (i.e. zero boys)?
We're choosing 3 girls from the 6 girls, or 6C3 = 6!/(3!)(3!) = 20.

So, there are 20 ways to form a 3-person group consisting of zero boys.
Subtracting 20 from 120, we get 100 ways of forming a 3-person group consisting of at least 1 boy.
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by tomada » Sat Feb 25, 2012 11:09 am
First I started with the total number of ways to choose a study group of 3 people from 10 people: 10C3 = 10!/(7!)(3!) = 120.

This number, 120, can be thought of as the sum of the following:

# of ways a 3-person group can be formed containing zero boys +
# of ways a 3-person group can be formed containing one boy +
# of ways a 3-person group can be formed containing two boys +
# of ways a 3-person group can be formed containing three boys

The above comprise an exhaustive set of possibilities which, as shown by 10C3, is 120.

The phrase at least one boy is represented by the lattermost three of the four possible scenarios shown above.

So, if we can calculate the first scenario - namely, the # of ways a 3-person group can be formed containing zero boys,
we can subtract that number from 120, and we'll have our answer.

How many ways are there to form a 3-person group consisting solely of girls (i.e. zero boys)?
We're choosing 3 girls from the 6 girls, or 6C3 = 6!/(3!)(3!) = 20.

So, there are 20 ways to form a 3-person group consisting of zero boys.
Subtracting 20 from 120, we get 100 ways of forming a 3-person group consisting of at least 1 boy.
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by tomada » Sat Feb 25, 2012 11:27 am
First I started with the total number of ways to choose a study group of 3 people from 10 people: 10C3 = 10!/(7!)(3!) = 120.

This number, 120, can be thought of as the sum of the following:

# of ways a 3-person group can be formed containing zero boys +
# of ways a 3-person group can be formed containing one boy +
# of ways a 3-person group can be formed containing two boys +
# of ways a 3-person group can be formed containing three boys

The above comprise an exhaustive set of possibilities which, as shown by 10C3, is 120.

The phrase at least one boy is represented by the lattermost three of the four possible scenarios shown above.

So, if we can calculate the first scenario - namely, the # of ways a 3-person group can be formed containing zero boys,
we can subtract that number from 120, and we'll have our answer.

How many ways are there to form a 3-person group consisting solely of girls (i.e. zero boys)?
We're choosing 3 girls from the 6 girls, or 6C3 = 6!/(3!)(3!) = 20.

So, there are 20 ways to form a 3-person group consisting of zero boys.
Subtracting 20 from 120, we get 100 ways of forming a 3-person group consisting of at least 1 boy.
I'm really old, but I'll never be too old to become more educated.
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