Permutations and Combinations help needed

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wadez0rz

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1) A box contains 6 pink marbles, 8 green marbles, and 3 red marbles. How many different samples of 5 can be selected if it has exactly 2 green marbles or exactly 2 red marbles?

2) How many 6 card hands have exactly 3 diamonds or exactly 3 black cards?

3) A box contains two red, four green, six black, and three blue balls. In how many ways can a sample of four balls be selected if At most two blue balls are drawn?


Thank you in advance.
 
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My apologies, thanks for the corrections!

The biggest problem I'm having has to do with, for example, the need for two green balls or two red balls in each sample in the first question, or the need for three diamonds or black cards. I know that the number of combinations is

C(n,r) = n!/(r!(n-r)!)

where n is the number of distinct objects and r is the number taken at a time.

For the first problem, I know that if you take out two green balls, there would be 6 left, but since you can only have two in one hand, does that mean that you really only have 9 balls left to use as your n? Likewise, for the red balls, you would have 14 left. I then figured that you would multiply the two combinations together, giving 252,252, but that seems unreasonably high.

I'm curious about that reasoning for the other two problems as well.

Does this seem right, or are my suspicions right?
 
wadez0rz said:
My apologies, thanks for the corrections!

The biggest problem I'm having has to do with, for example, the need for two green balls or two red balls in each sample in the first question, or the need for three diamonds or black cards. I know that the number of combinations is

C(n,r) = n!/(r!(n-r)!)

where n is the number of distinct objects and r is the number taken at a time.

For the first problem, I know that if you take out two green balls, there would be 6 left, but since you can only have two in one hand, does that mean that you really only have 9 balls left to use as your n? Likewise, for the red balls, you would have 14 left. I then figured that you would multiply the two combinations together, giving 252,252, but that seems unreasonably high. It certainly is too high.
How many ways in total can you select 5 balls from 17. The answer is
17! / [(17 - 5)! * 5!] = 17 * 16 * 15 * 14 * 13 / (5 * 4 * 3 * 2) =
17 * 16 * 15 * 14 * 13 / (15 * 8) = 17 * 2 * 1 * 14 * 13 = 6,188, way less than 252,252

Imagine the balls have numbers on them. So green balls are numbered 1 through 8, pink balls are numbered 1 through 6, and red balls are numbered 1 through 3.

How many ways can you select two green balls from eight. You do not care about order; picking green 3 first and green 5 second is the same to you as picking green 5 first and green 3 second. So the answer is 8! / [(8 - 2)! * 2!] = 8! / (6! * 2!) = 7 * 8 / 2 = 28.
How many ways can you select two pink balls from six. By the same logic, 6! /[(6 - 2!) * 2!] = 6! / (4! * 2!) = 6 * 5 / 2 =15.
How many ways can you select one red ball from 3? Obviously 3! / [(3 - 1)! * 1!] = 3! / (2! * 1!) = 3 / 1 = 3.
So, for each possibility of pink balls, you have three possibilites for red balls, giving 3 * 15 = 45 possibilities for pink and red balls.
And for each possibility of red and pink balls, you have 28 possibilities for green balls, giving 45 * 28 = 1260.

I'm curious about that reasoning for the other two problems as well.

Does this seem right, or are my suspicions right?
Click to expand...
Now give the next two problems another look.

EDIT: PKA may be right about what the question is asking for. What is certain is that my analysis above is for the question of how many ways can you select a sample that has exactly two red balls AND exactly two green balls. It is not the correct analysis for the question of how many ways to select a sample that has exactly two red balls OR exactly two green balls. If that is the question, you must determine how many ways to select exactly two red balls, how many ways to select exactly two green balls, and how many ways to select exactly two red balls and exactly two green balls, and combine those three numbers properly. That is, the analysis I gave does less than one third of the the problem. That was careless of me, and I apologize. Moreover, as PKA points out, the question may be entirely different.
 
Last edited:
wadez0rz said:
1) A box contains 6 pink marbles, 8 green marbles, and 3 red marbles. How many different samples of 5 can be selected if it has exactly 2 green marbles or exactly 2 red marbles?
Click to expand...
I think that the answer to this question is seven.
\(\displaystyle \begin{array}{*{20}c} P & G & R \\
\hline
0 & 2 & 3 \\
0 & 3 & 2 \\
1 & 2 & 2 \\
2 & 2 & 1 \\
2 & 1 & 2 \\
3 & 2 & 0 \\
3 & 0 & 2 \\
\end{array}\)
The question asks "How many different samples of 5"
Even if the balls were numbered it does not say "How many different ways can a sample of 5 be chosen?"
 
The exercise states "or".

I agree with pka.

If a collection of five marbles contains at least 2 green marbles, that's sufficient. Those two green marbles may be any pair from the set of eight (i.e., we don't count all of the different green pairs), and the colors of the remaining three marbles in the collection do not matter.

If a collection of five marbles contains at least 2 red marbles, that's also sufficient. Again, any of the three available red marbles may comprise that selected pair, and the colors of the remaining three marbles in the collection do not matter.

List all of the possible combinations with 2 green. List all of the possible combinations with 2 red. Eliminate any duplicate(s) between the lists.

Cheers :cool:
 
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