1. ## combinatorial question

Here is my question, I'm very confused. I struggle with word problems. Could someone please point me in the right direction?

A store has 10 identical boxed stereos to store on 3 shelves. How many ways can the boxes be
stored on the shelves if:

(a) there are no restrictions on the number of stereos on each shelf?
(b) the top shelf must have exactly 2 stereos on it?
(c) each shelf must have at least 2 stereos on it?
(d) no shelf can have more than 5 stereos on it?

Greatly Appreciated.

hmm for question (a)

is 3x 10! correct?

2. I think... after much confusion and struggling, that I have it for (a)

is it: 12/ (10!)(2!) = 66?

(b) 9!/(8!)(1!) = 9

(c) 8!/(6!)(2!) = 28

(d) (12/ (10!)(2!)) - (6!/(4!)(2!)) = 51

this is what I got... *phew* I would appreciate it if someone could tell me if I did this right =]

3. yes! for (a) and (b)

would you explain (c), (d)

PS: plz refer to http://www.mathhelpforum.com/math-he...-question.html

4. oh wait... (c) should be... (6!) / (4!)(2!)

and now im not sure at all what I did for (d) =p help anyone?

5. Originally Posted by yoonsi
oh wait... (c) should be... (6!) / (4!)(2!)

and now im not sure at all what I did for (d) =p help anyone?
Hi - (c) yes

(d) first find how many ways are there when at least one of the shelf has > 5 boxes (which is complement of what is asked in the question)

1)observe there can be only 1 shelf with >5 boxes (this observation helps us greatly otherwise we would have to do something more complicated) - so in how many ways can you select the shelf with>5 boxes?

2)once you are done with (1), put 6 boxes in the selected shelf (well it should at least have 6 (>5) boxes as per our argument!) - rest 4 can be distributed in anyways you like among the 3 shelves. How many? You have already solved this bit

Combine (1) and (2) to find how many ways are there when at least one of the shelf has > 5 boxes

Now that you have this ans - answering the original question should be straight fwd

ill give you the final answer - 21 (if i'm not wrong in my calcs)

Thanks

6. Thanks so much,

what I did was:
(d) (12/ (10!)(2!)) - (6!/(4!)(2!)) = 51

when I should have:
(d) (12/ (10!)(2!)) - 3*(6!/(4!)(2!)) = 21

Thanks =D