Combinatorics pencil box problem

[Roger.Robert]

In a pencil box there are 8 crayons:3 red,2 blue and 3 black.If we take 5 crayons from the pencil box, how many possibilities are there so that at least two crayons are red?

 

[JeffM]

In a pencil box there are 8 crayons:3 red,2 blue and 3 black.If we take 5 crayons from the pencil box, how many possibilities are there so that at least two crayons are red?

You should know by now that we want to see what work you have done or at least hear what ideas you have had.

 

[Steven G]

In a pencil box there are 8 crayons:3 red,2 blue and 3 black.If we take 5 crayons from the pencil box, how many possibilities are there so that at least two crayons are red?

If you took at least two red crayons, then how many red crayons might you have taken? Show us your work!

 

[Roger.Robert]

I think the final answer is 6.If I choose 2 red pencils I can make 3 combinations with the rest of the pencils.If I choose 3 red pencils I also can choose 3 combinations.So in total 6.
But I was wondering what these types of problems are called(where there are a limited number of identical objects,and count the number of combinations after picking n number of objects), so I can find the general methods to solve.

 

[Denis]

In a pencil box there are 8 crayons:3 red,2 blue and 3 black.If we take 5 crayons from the pencil box, how many possibilities are there so that at least two crayons are red?

That's badly worded. Should be:
if 5 crayons are taken AT RANDOM,
what is the PROBABILITY that at least 2 crayons are red?

Are you a student attending math classes?

 

[Dr.Peterson]

That's badly worded. Should be:
if 5 crayons are taken AT RANDOM,
what is the PROBABILITY that at least 2 crayons are red?

Are you a student attending math classes?

No, I think this is a problem about combinatorics, not probability.

What is badly worded is that it is not clear what counts as a different "possibility". Are the crayons considered distinguishable, or are we only considering how many of each color there are? (If it is ultimately going to be a probability question, then we have to treat them as distinguishable, in order to have equally likely outcomes.) Also, does order matter?

 

[Dr.Peterson]

I think the final answer is 6.If I choose 2 red pencils I can make 3 combinations with the rest of the pencils.If I choose 3 red pencils I also can choose 3 combinations.So in total 6.
But I was wondering what these types of problems are called(where there are a limited number of identical objects,and count the number of combinations after picking n number of objects), so I can find the general methods to solve.

I wouldn't consider this, as it stands, to be an example of any particular type of problem for which there would be a formula to learn. These problems of combinatorics require careful thinking on a case-by-case basis, though there may be formulas to be used for parts of them.

As it stands, I can't be sure how to approach it, because the meaning of the problem as given is not clear. If this is a problem you were given, please quote the problem exactly as given, including any general instructions in the context.

You say here that the pencils are considered to be identical (indistinguishable apart from color), so that perhaps the contents of the box can be represented as RRRBBKKK. I think you are ignoring order, so you want the number of combinations of 5 elements of this multiset in which two are R: R R _ _ _. (Neither of the two assumptions you have used is stated in the problem.)

Therefore, the question is equivalent to asking for the number of combinations of RBBKKK taken 3 at a time. There are formulas for problems similar to this (such as counting permutations of a multiset, or all multisets of a given set), but I think this one just has to be counted carefully.

Let's see if your count is correct. I think you are saying that if the remaining 3 pencils include no R, they can be BBK, BKK, or KKK; and if they include one R, they can be RBB, RBK, or RKK. Yes, this looks correct. And that may be exactly the way to solve it.

 

[pka]

I think the final answer is 6.If I choose 2 red pencils I can make 3 combinations with the rest of the pencils.If I choose 3 red pencils I also can choose 3 combinations.So in total 6.
But I was wondering what these types of problems are called(where there are a limited number of identical objects,and count the number of combinations after picking n number of objects), so I can find the general methods to solve.

Here is a completely different way to approach this problem.
Expand \(\displaystyle \underbrace {({x^2} + {x^3})}_{red}\underbrace {(1 + x + {x^2})}_{black}\underbrace {(1 + x + {x^2} + {x^3})}_{blue}\) See we take a minimum of two reds, a min of zero black and max of two, etc.
The expansion is \(\displaystyle x^2+3x^3+5x^4+6x^5+5x^6+3x^7+x^8\) see here.

The term \(\displaystyle 6x^5\) tells us the there are six ways to choose five pencils including two reds.
You are indeed correct about the 6.