Problem Solving

I really don't know how to solve this one!

A farmer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

OA: 100

rest assured, farmer would make three selection sets with one black, two black and three black sheep, because of at Least One black condition. So we fill the combination sets the following way/group
4 black * 6 white * 5 white = 120
4 black * 3 black * 6 white = 72
4 black * 3 black * 2 black = 120
total makes 216 ways/groups

garuhape wrote:I really don't know how to solve this one!

A farmer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

OA: 100

A farmer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

(4c1*6c2)+(4c2*6c1)+(4c3)
= 60+36+4 = 100

your solution is correct; I had to use combined sets not permutation --> black by white.

HSPA wrote:A farmer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

(4c1*6c2)+(4c2*6c1)+(4c3)
= 60+36+4 = 100

HSPA wrote:A farmer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

(4c1*6c2)+(4c2*6c1)+(4c3)
= 60+36+4 = 100

What does c1, c2 and c3 refer to? Thx

C ist fuer combinations....
du kennst combinations oderrrrr???
Bitte lernen auf deiser link https://www.mathwords.com/c/combination_formula.htm

hi HSPA, wie geht's
deine antworte ist richtig.
danke schon

HSPA wrote:C ist fuer combinations....
du kennst combinations oderrrrr???
Bitte lernen auf deiser link https://www.mathwords.com/c/combination_formula.htm

no black in 6c3 = !6/!3!3 = 20
Req no = 10c3 - 20 = 10*9*8*7/3*2 - 20 = 260

garuhape wrote:I really don't know how to solve this one!

A farmer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

OA: 100

garuhape wrote:I really don't know how to solve this one!

A farmer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

OA: 100

Good combinations = Total Combinations - Bad Combinations

In the problem above:
Good combinations = number of combinations of 3 that include at least 1 black sheep
Total combinations = number of combinations of 3 that can be formed from the 10 sheep
Bad combinations = number of combinations of 3 that include no black sheep = number of combinations of 3 that can be formed from the 6 white sheep.

Total combinations = 10C3 = 120.
Bad combinations = 6C3 = 20.

Thus, good combinations = 120-20 = 100.

best way to do this mitch...

GMATGuruNY wrote:

garuhape wrote:I really don't know how to solve this one!

A farmer has 10 sheep 4 black and 6 white. He would like to take 3 of the sheep to get shaved. How many different groups of 3 sheep can he select that would have at least one black one?

OA: 100

Good combinations = Total Combinations - Bad Combinations

In the problem above:
Good combinations = number of combinations of 3 that include at least 1 black sheep
Total combinations = number of combinations of 3 that can be formed from the 10 sheep
Bad combinations = number of combinations of 3 that include no black sheep = number of combinations of 3 that can be formed from the 6 white sheep.

Total combinations = 10C3 = 120.
Bad combinations = 6C3 = 20.

Thus, good combinations = 120-20 = 100.

That's a really nice explanation. Thanks!!!

I just wanted to apply what I have learned on this particular problem:

A manager needs to hire 3 people to work in her marketing department, and one of which is to be the team leader. If there are 10 potential candidates, how many distinct ways can she fill these three positions?


In this case, there are only good combination and no bad ones.

10!/(3!*7!) = 720/6 = 120

However, the answer is 360. So I guess that I have to multiply the 120 by 3. But why? Aren't the 120 already all the possible combination?

Thx

garuhape wrote:I just wanted to apply what I have learned on this particular problem:

A manager needs to hire 3 people to work in her marketing department, and one of which is to be the team leader. If there are 10 potential candidates, how many distinct ways can she fill these three positions?


In this case, there are only good combination and no bad ones.

10!/(3!*7!) = 720/6 = 120

However, the answer is 360. So I guess that I have to multiply the 120 by 3. But why? Aren't the 120 already all the possible combination?

Thx

So now the manager has selected the 3 candidates in 10C3 ways = 120 ways. That is fine.
However, he wants to make one of them the team lead.
You have 3 candidates now, and want one to make team lead. How many ways can you do it : choose one out of 3 : right ?
That means : 3C1 = 3 ways
So, total no. of ways = (No. of ways to select 3 candidates out of 10) * (No. of ways to select 1 candidate as team lead from the 3 selected candidates ) = 120*3 = 360.

Hope that helps !

garuhape wrote:I just wanted to apply what I have learned on this particular problem:

A manager needs to hire 3 people to work in her marketing department, and one of which is to be the team leader. If there are 10 potential candidates, how many distinct ways can she fill these three positions?


In this case, there are only good combination and no bad ones.

10!/(3!*7!) = 720/6 = 120

However, the answer is 360. So I guess that I have to multiply the 120 by 3. But why? Aren't the 120 already all the possible combination?

Thx

Start with the most restricted position, which is the team leader.
Number of choices for team leader = 10.
A combination of 2 people must be put together with the team leader.
Number of combinations of 2 that can be formed from the 9 remaining people = 9C2 = 36.
To combine the number of choices we have for the team leader with the number of choices we have for the other two positions, we multiply the results above:
10*36 = 360.

If we simply count the number of combinations of 3 that can be formed from the 10 candidates, we'll be undercounting the total number of possible assignments.
ABC is one combination of 3 people, but Candidate A as team leader with B and C his associates is a different assignment from Candidate B as team leader with A and C his associates.
Thus, to account for all the different ways that the team leader can be assigned, we first we need to count how many choices we have for the team leader -- the most restricted position -- and then proceed from there.