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Expert replies
by VirusWar » Thu Jan 12, 2017 1:49 pm
Hello, I have a math assignment that needs solving, and I need a bit of help.
It goes like this:
You have two groups, 6 boys and 5 girls, and you need to make a group of three with at least one girl. How many combinations are there?
I get that there are 3 types of groups, 'one girl - two boys', 'two girls - one boy' and 'three girls'.
The one with three girls I solved, its 5c3 = 10.
But when I have two groups I don't know how its done, so I need a bit of explaining and it would be great if there is an equation.

Thanks in advance!
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Source: — Problem Solving |

by DavidG@VeritasPrep » Thu Jan 12, 2017 1:54 pm
VirusWar wrote:Hello, I have a math assignment that needs solving, and I need a bit of help.
It goes like this:
You have two groups, 6 boys and 5 girls, and you need to make a group of three with at least one girl. How many combinations are there?
I get that there are 3 types of groups, 'one girl - two boys', 'two girls - one boy' and 'three girls'.
The one with three girls I solved, its 5c3 = 10.
But when I have two groups I don't know how its done, so I need a bit of explaining and it would be great if there is an equation.

Thanks in advance!
One method: find the number of total possibilities and then subtract out the undesired possibilities.

1) Total ways that we can select a group of 3 from 11 people - 11C3 = (11*10*9)/(3*2) = 11*5*3 = 165.

2) Find the number of undesired outcomes: In this case, if we want at least one girl, our undesired outcome is no girls, or all boys. So how many ways can we select 3 boys from 6? 6C3 = (6*5*4)/(3*2) = 20

3) Total - undesired = 165 - 20 = 145
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by DavidG@VeritasPrep » Thu Jan 12, 2017 2:12 pm
VirusWar wrote:Hello, I have a math assignment that needs solving, and I need a bit of help.
It goes like this:
You have two groups, 6 boys and 5 girls, and you need to make a group of three with at least one girl. How many combinations are there?
I get that there are 3 types of groups, 'one girl - two boys', 'two girls - one boy' and 'three girls'.
The one with three girls I solved, its 5c3 = 10.
But when I have two groups I don't know how its done, so I need a bit of explaining and it would be great if there is an equation.

Thanks in advance!
Alternatively, we can work through each desired scenario.

1) one girl and two boys
Number of ways we can select one girl from a pool of five: 5
Number of ways we can select two boys from a pool of 6: 6C2 = 6*5/2 = 15
Now just multiply them. Number of ways we can select one girl and two boys = 5 * 15 = 75

2) Two girls and one boy
Number of ways we can select two girls from a pool of five: 5C2 = 5*4/2 = 10
Number of ways we can select one boys from a pool of 6 = 6
Number of ways we can select two girls and two boy = 10 * 6 = 60

3) Three girls
Number of ways we can select three girls from a pool of five: 5C3 = 5*4*3/3*2 = 10

Add up the three scenarios: 75 + 60 + 10 = 145
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by [email protected] » Thu Jan 12, 2017 2:22 pm
Hi VirusWar,

David has shown two great ways to approach this question, so I won't rehash any of that here. Instead, I want to ask about your focus - are you actually studying for the GMAT or are you studying for a 'math class?' I ask because GMAT questions are designed in a rather specific way (and they include 5 answer choices), so if you're going to post GMAT questions then you should make sure to post the FULL prompt. If you're just studying for a math class though, you will likely be asked to deal with concepts (and answer questions) that don't align with what the actual GMAT focuses on - and posting those questions would not be beneficial for site users who are actually here training to face the GMAT.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Jeff@TargetTestPrep » Mon Jan 16, 2017 5:11 pm
VirusWar wrote:Hello, I have a math assignment that needs solving, and I need a bit of help.
It goes like this:
You have two groups, 6 boys and 5 girls, and you need to make a group of three with at least one girl. How many combinations are there?
I get that there are 3 types of groups, 'one girl - two boys', 'two girls - one boy' and 'three girls'.
The one with three girls I solved, its 5c3 = 10.
But when I have two groups I don't know how its done, so I need a bit of explaining and it would be great if there is an equation.

Thanks in advance!
In "at least one" problems, we can employ the following formula:

Total ways to select a group of 3 people = (# ways with 0 girls and 3 boys) + (# ways with 1 girl and 2 boys) + (# ways with 2 girls and 1 boy) + (# ways with 3 girls and 0 boys)

Thus, the number of ways to select at least one girl is equal to:

(Total ways to select a group of 3 people) - (# ways with 0 girls and 3 boys)

Since there are 6 boys and 5 girls, the total number of ways to select a group of 3 people is 11C3:

11C3 = 11!/3!8! = (11 x 10 x 9)/3! = (11 x 10 x 9)/(3 x 2 x 1) = 11 x 5 x 3 = 165

The number of ways to select 0 girls and 3 boys is 5C0 x 6C3:

5C0 = 1 (recall that nC0 is always equal to 1)

6C3 = 6!/3!3! = (6 x 5 x 4)/3! = (6 x 5 x 4)/(3 x 2 x 1) = 5 x 4 = 20

The number of ways to select a group of three people with NO girls is 20, and the total number of ways to select a group of three people is 165. Thus, the number of ways to select a group of three people with AT LEAST one girl is 165 - 20 = 145.

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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by VirusWar » Fri Jan 20, 2017 11:34 am
Thanks a lot for your help, appreciate it!
Cheers!
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by Matt@VeritasPrep » Wed Feb 01, 2017 5:16 pm
Since your teacher will probably like a quick and snappy solution, how about this one:

# of groups with at least one girl =

Total - (# of groups with no girls) =

(11 choose 3) - (6 choose 3) =

11!/(8!3!) - 6!/(3!3!) =

(1/3!) * (11!/8! - 6!/3!) =

1/6 * (11*10*9 - 6*5*4) =

145
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