Permutation/Combination problem

[shamch01]

Please can somebody help solving this one? Related to permutation - combination problem I suppose. Hope this is the right category to post the same. Thanks in advance.

 

[Steven G]

This is the correct category to place your question in. But according to the forum guideline you are requested to post your question in the correct category AND show your attempt to do this problem. This forum is a help forum, not a submit your problem and receive an answer forum.

So please post your attempt at doing this problem and then some helper will point you in the correct direction to help YOU arrive at the correct answer.

 

[MarkFL]

Hello, and welcome to FMH!

I would approach this on a case by case basis:

1.) How many ways can she put all 3 straws into 1 cup?

2.) How many ways can she put 2 straws in one cup and one straw in another?

3.) How many ways can she put one straw on one cup?

 

[Harry_the_cat]

How many ways can she put all 3 straws in one cup?
How many ways can sge put exactly 1 straw in each cup?
How many ways can she put exactly 2 straws in one cup, and one in another?
Are tgere any other options?

 

[Harry_the_cat]

Snap MarkFL. Great minds ...

 

[Steven G]

I like to think about these type problems as follows.

Consider three vertical lines, | | |. Let these line represent the cups.

You want to place 3 x's into the cups.

Represent which cup gets the x by putting the x to the left of the line.

So xx| | x| means 2 x's in the 1st cup, no x's in the 2nd cup and one x in the third cup.

Note that it makes no sense to have an x after the third line

So this problem reduces to how many ways can we arrange three x's and three |'s with no x being at the end.

Now this is a combinatoric/permutation problem.

The method provided above is the way to do this one problem because the numbers are small but with larger numbers you should use my method.

In my opinion it is better to learn the method that works for all cases.

So gives this method a try and let us know how it worked out for you!

By the way, this was the method which I used on my very first post on this forum and our friend Denis wondered who I was.

 

[pka]

The number of ways to place $N$ identical objects into $K$ distinct cells is $\dfrac{[N+(K-1)]!}{N!(K-1)!}$

 

[shamch01]

Hello, and welcome to FMH!

I would approach this on a case by case basis:

1.) How many ways can she put all 3 straws into 1 cup?

2.) How many ways can she put 2 straws in one cup and one straw in another?

3.) How many ways can she put one straw on one cup?

Thanks MarkFL.

Feel great to be part of this forum.

I have worked out this problem in a similar way.

However, I couldn't come to the answer which is 10.


1.) How many ways can she put all 3 straws into 1 cup? - thats should be 3 different ways

2.) How many ways can she put 2 straws in one cup and one straw in another? - thats also 3 different ways

3.) How many ways can she put one straw on one cup? - can be 1 way, as the straws are identical

Am I correct/logical in thinking?

The correct answer is 10.

I was adding up the ways : 1+3+3 = 7

However, it should be 1+ 3×3 = 10 - couldn't follow why should it be multiplied

Thanks

 

[Romsek]

1.) How many ways can she put all 3 straws into 1 cup? - thats should be 3 different ways

2.) How many ways can she put 2 straws in one cup and one straw in another? - thats also 3 different ways

3.) How many ways can she put one straw on one cup? - can be 1 way, as the straws are identical

Am I correct/logical in thinking?

The correct answer is 10.

I was adding up the ways : 1+3+3 = 7

However, it should be 1+ 3×3 = 10 - couldn't follow why should it be multiplied

Thanks

1) is correct

2) is incorrect. Pick a cup to put 2 straws into. That's 3 choices. Now pick one of the remaining cups to put the last straw into.
That's another 2 choices. Thus there are 3x2=6 ways to put 2 straws into 1 cup and 1 into another.

3) is correct

3 + 6 + 1 = 10

 

[Steven G]

Thanks MarkFL.

Feel great to be part of this forum.

I have worked out this problem in a similar way.

However, I couldn't come to the answer which is 10.


1.) How many ways can she put all 3 straws into 1 cup? - thats should be 3 different ways

2.) How many ways can she put 2 straws in one cup and one straw in another? - thats also 3 different ways

3.) How many ways can she put one straw on one cup? - can be 1 way, as the straws are identical

Am I correct/logical in thinking?

The correct answer is 10.

I was adding up the ways : 1+3+3 = 7

However, it should be 1+ 3×3 = 10 - couldn't follow why should it be multiplied

Thanks

2.) How many ways can she put 2 straws in one cup and one straw in another? - thats also 3 different ways----Not Correct
Remember that you can switch where you placed the 2 straws and the 1 straw. This will yield 6.
You were told the answer was 10, a small number, so why not try to list them? Then you would have seen where the extra three comes from.

 

[shamch01]

1) is correct

2) is incorrect. Pick a cup to put 2 straws into. That's 3 choices. Now pick one of the remaining cups to put the last straw into.
That's another 2 choices. Thus there are 3x2=6 ways to put 2 straws into 1 cup and 1 into another.

3) is correct

3 + 6 + 1 = 10

Thank you so much Romsek.

Thats where I was wrong!!!