Disregard my previous post, the solution is correct:
Now label the four (distinct) fingers A,B,C and D, and the nine (dsitinct) rings 1,2,3,...,9.
What we are asked to do is find the total number of ways in which we can place the 9 rings on the four fingers, if the order of the rings on each finger is considered, i.e. is taken into account. So, to clear up anya mbiguity, if all 9 rings are placed on finger A, then being placed as 123456789 is different (i.e. is counted seperately to) the fingers being placed 987654321, or 135792468, etc.
To approach this, i will first consider a similar, yet different problem:
First consider if the question asked for the number of combinations, and NOT permutations.
In this problem, if all 9 rings are on finger A, no matter what order they are in, they are counted as the same combination. So to find the number of combinations, consider the following 'table' (denote each ring as * since they are identical for our purposes, since we are only considering combinations):
#...Finger A........Finger B..........Finger C ........Finger D
--------------------------------------------------------------
1. *********
2. ********........*
3. ******* ...................................* ................ *
4. ***................*** ..................... ** ............. *
[I have used '......' to space out the *'s]
etc...
[I have only listed 4 cases - there are in fact 220 ,as we will see below]
We want to find out how many rows are in this complete table. Consider instead of table form, the diagrams for the examples listed in the table:
*********|||
********|*||
*******||*|*
***|***|**|*
That is, '|' is a barrier, seperating the rings on finger A, from that of finger B, C, and D.
Clearly we have 9 (identical) rings (*'s) and 3 barriers (|'s - 3 barriers will speerate the rings into the four fingers). Thuis the number of combinations of these are:
12!/(3!9!) ----> Since there are 12 items we are arranging in a straight line, 3 of which are | and are identical and 9 of which are * and are identical, so this problem is equivalen to saying 'how many words can be made from the letters XXXXXXXXXYYY'.
This is equal to 12C3, or 12C9 = 220 [There is a general formula for this, but i wont get into that].
Now this is not our original problem. What i will proceed to show is that for each of these 12C9 combinations, we can arrange the rings (which are not identical in our original problem, but are distinct) in 9! ways, and so if we want the number of permutations all we do is multiply 12C9 by 9!.
Consider some combinations:
1) for the case where all 9 rings go on one finger, yes there are 9! ways of arranging the rings on that finger. So for 4 of the above 12C9=220 combinations (since we can put all 9 rings on any of 4 fingers), yes we should indeed multiply by 9! to get the corresponding permutations.
2) for the case where we put 2 rings on Finger A, 2 rings on Finger B, 2 rings on Finger C, and 3 rings on Finger D, there ARE 9! ways to arrange the rings, but lest us see why; there are 9P2*7P2*5P2*3P3=362,880 [Note that 9!=362,880] ways, since we must place 2 of the 9 rings on finger A, order is important, then 2 of the reaining 7 rings on finger B, order is important, 2 of the remaining 5 rings on finger C, order is important, and 3 of the remaining 3 rings on finger D, oder is important [Note 3P3=3!, if you want to think about it that way]. So in this case, there is also 9! ways, but notice how we got the answer - using P's.
3) for the case where we put 1 ring on finger A nad 8 on finger B, there are 9P1*8P8=362,880 [=9!] ways, using a similar argument to 2) above.
We can repeat this for each of the 220 cases, but we don't have to,w e know each case will have 9! ways of arranging the rings.
So the required answer is 12C3*9!, as given.
Note: a spreadsheet program (like excel) is useful if you are going to list out all the cases (in fact the 220 cases can be further reduced into 18 cases, which we can then get the 220 from, but i won't get into that).
So as a simple answer:
12C3 ways to arrange the rings onto the fingers [above diagram is useful to show this], and we multiply by 9! since each ring is distinct.
