Q on combinatorics.

[Sonal7]

I found part (a) fair doable.There are three possible combinations that will give 7 coconuts, 223, 323, or 322 and therefore its the addition of the different combinations from each tree 3C2*4C2*5C3 + 3C3*4C2*5C2 +3C2*3C3*5C2.

I am unable to understand the answer to part (b). I thought all the different arrangements might be 4!*5! %3!2!

 

[pka]

I am unable to understand the answer to part (b). I thought all the different arrangements might be 4!*5! %3!2!

Think of the three from tree one as a single box. The box's content can be arranged in \(3!\) ways.
The box and the other nine coconuts comprise ten items.
The the total number of ways to arrange is \(3!\cdot 10!\). Do you see why?

 

[Sonal7]

Think of the three from tree one as a single box. The box's content can be arranged in \(3!\) ways.
The box and the other nine coconuts comprise ten items.
The the total number of ways to arrange is \(3!\cdot 10!\). Do you see why?

I have been thinking that the 3 coconuts are not arranged differently and they are from the same tree. So we are talking about permutations, C1, C2 and C3 being three coconuts from the first tree. If they look the same then the implication is that which ever way you arrange it would not matter. Where did you get 10! from? It matches the right answer.

 

[Dr.Peterson]

What are the answers you are given? Is your answer to part (a) "correct"?

If so, then they are treating the individual coconuts as distinguishable (as you did), and pka's answer would also be correct. (The 10 comes from thinking of the coconuts as 1 group from the first tree, 4 nuts from the second, and 5 from the third, and arranging those 10 items in any order.)

I would have taken the wording of the problem as (possibly) suggesting that it is only the origin of each coconut that can be identified, so that individual coconuts are not distinguishable, so the problem would be equivalent to arranging AAABBBBCCCCC, keeping the AAA together. But it sounds like that is not what was intended.

 

[Sonal7]

What are the answers you are given? Is your answer to part (a) "correct"?

If so, then they are treating the individual coconuts as distinguishable (as you did), and pka's answer would also be correct. (The 10 comes from thinking of the coconuts as 1 group from the first tree, 4 nuts from the second, and 5 from the third, and arranging those 10 items in any order.)

I would have taken the wording of the problem as (possibly) suggesting that it is only the origin of each coconut that can be identified, so that individual coconuts are not distinguishable, so the problem would be equivalent to arranging AAABBBBCCCCC, keeping the AAA together. But it sounds like that is not what was intended.

I need to think about it for a few minutes. Like where the 3! comes from.

 

[pka]

But it sounds like that is not what was intended.

It does specify that "The coconuts from each tree can be identified." I would have though that adding "as being from a particular tree but were otherwise identical." would cleared any misunderstanding. Then the \(AAABBBBCCCCC\) model works.
Keeping the A's together, that string can be arranged in \(\dfrac{(10!)}{(4!)(5!)}\) ways.

 

[Sonal7]

It does specify that "The coconuts from each tree can be identified." I would have though that adding "as being from a particular tree but were otherwise identical." would cleared any misunderstanding. Then the \(AAABBBBCCCCC\) model works.
Keeping the A's together, that string can be arranged in \(\dfrac{(10!)}{(4!)(5!)}\) ways.

that gives 1260 and 3! *10! which is the books answer gives a horrendously longer number. I thought pka, your original answer was the correct one. Why did you change it? Will you please explain the previous answer

 

[Sonal7]

Think of the three from tree one as a single box. The box's content can be arranged in \(3!\) ways.
The box and the other nine coconuts comprise ten items.
The the total number of ways to arrange is \(3!\cdot 10!\). Do you see why?

This is the correct answer but I am not clear how you derived it. I think wording is the problem.

 

[Sonal7]

I got it now, I think its not worded well. but they are asking for arrangements not different combinations. I found the part (b) pretty hard.

 

[Dr.Peterson]

I think pka explained it quite clearly. Put the three coconuts into a bucket, and you can arrange the bucket and 9 coconuts in 10! ways. Then take the three out of the bucket and put them in a line where the bucket was, which you can do in 3! ways. Then multiply.

And clearly the author does intend the coconuts to be individually distinguishable, though the wording was a little ambiguous.

 

[Sonal7]

Thank you very much. It a bit of pain. They said arrangements and that is permutations technically speaking in mathematical language. Thank you for clarifying this.