Can Someone tell me how to solve this perms/combs question :D (1 Viewer)

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How many ways are there to place nine different rings on the
four fingers of your right hand (excluding the thumb) if:
(i) the order of the rings on a finger does not matter?
(ii) the order of the rings on a finger is considered?

I got the first part, but fail to understand the 2nd. Explain please :D

ps. I realised something:

I forgot to mension. The question is worded badly. It means that you have to use ALL nine rings on the hand. i.e. the first part is 4^9. I didnt realise this till i looked at the answer, but the answersdoesnt say why the 2nd part is as it is.
 
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Flying_Vagina

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9P4. It's a permutation as order matters. part2 differs from 1 in...
Lets say there are 9 rings called A,B,C,D,E,F,G,H,I. in part 1, ABCD, BCDA, CDAB, etc etc, are all considered as 1 way of arranging them as order doesnt matter, and they are all the same rings. BUT, in part 2, ABCD, BCDA, CBDC etc etc are considerd as 3 different ways, as they are in different orders. Hence why 9p4 is greater than 9c4.
 

hscishard

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How many ways are there to place nine different rings on the
four fingers of your right hand (excluding the thumb) if:
(i) the order of the rings on a finger does not matter?
(ii) the order of the rings on a finger is considered?

I got the first part, but fail to understand the 2nd. Explain please :D
First one is comb
Second one is Perm

Second should be greater than first.
 

CheesePlease

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9P4. It's a permutation as order matters. part2 differs from 1 in...
Lets say there are 9 rings called A,B,C,D,E,F,G,H,I. in part 1, ABCD, BCDA, CDAB, etc etc, are all considered as 1 way of arranging them as order doesnt matter, and they are all the same rings. BUT, in part 2, ABCD, BCDA, CBDC etc etc are considerd as 3 different ways, as they are in different orders. Hence why 9p4 is greater than 9c4.
This doesn't seem right to me. Obviously it is a question of permutations, but 9P4 could only be right if you interpret the question as "here's 9 rings, how many differently ordered groups of 4 can you have?"

I read that question more as "here's 9 rings, place them all on your 4 fingers, i.e. more than one ring can go on each finger". So if you consider the 4 fingers A, B, C & D, one of the combinations would be all 9 rings on A and none on B, C & D. Another combination 8 rings on A, 1 on B and none on C & D.

This would yield a vastly different number of combinations and permutations wouldn't it?

Sleight, you say you solved part 1, how did you do it and what was the answer? This may help us realise the perm/comb difference the question is testing. Is it from a textbook?
 

Flying_Vagina

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This doesn't seem right to me. Obviously it is a question of permutations, but 9P4 could only be right if you interpret the question as "here's 9 rings, how many differently ordered groups of 4 can you have?"

I read that question more as "here's 9 rings, place them all on your 4 fingers, i.e. more than one ring can go on each finger". So if you consider the 4 fingers A, B, C & D, one of the combinations would be all 9 rings on A and none on B, C & D. Another combination 8 rings on A, 1 on B and none on C & D.

This would yield a vastly different number of combinations and permutations wouldn't it?

Sleight, you say you solved part 1, how did you do it and what was the answer? This may help us realise the perm/comb difference the question is testing. Is it from a textbook?
I think your thinking about it too much, if it was what you just said, it would be too inaccurate, as everyone's finger sizes are different lol. And who the fuck wears more than 1 ring on a finger. But it is plausible to think what you just thought. In a hsc exam, they would have in brackets (This person only puts on one ring on each finger')
 

CheesePlease

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Yeah I guess so. I wouldn't really know how to solve it if it is the question I thought, well it would take about 20 times longer if wanted to do it the long way anyway. In an Exam I sure hope they'd be more specific.

Well if the answer to part 1 is 9C4 = 126, then part 2 is 9P4 = 3024.

And if not then we may have a fun question on our hands.
 

lychnobity

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1) 9C4
2) 9P4

First part requires picking any random 4, doesn't matter where each ring goes

Second part wants specific ORDERS.
 

lychnobity

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erm, why would part 2 be greater than part 1? There are more restrictions for the 2nd one...
You're wrong.

Part i basically wants you to choose any 4 out of 9 rings. Part ii wants 4 rings to be chosen, AND THEN wants a finger to be chosen for each ring; this is not a restriction, it's another option (crap wording).

eg. Q W E R T Y U I O are the ring's names

for part i) QWER and REWQ would be the same, but part ii) would treat them as 2 different cases

If you still aren't convinced, consult some textbook.
 
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I forgot to mension. The question is worded badly. It means that you have to use ALL nine rings on the hand. i.e. the first part is 4^9. I didnt realise this till i looked at the answer, but the answersdoesnt say why the 2nd part is as it is.
 

beardedwoman

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You're wrong.

Part i basically wants you to choose any 4 out of 9 rings. Part ii wants 4 rings to be chosen, AND THEN wants a finger to be chosen for each ring; this is not a restriction, it's another option (crap wording).

eg. Q W E R T Y U I O are the ring's names

for part i) QWER and REWQ would be the same, but part ii) would treat them as 2 different cases

If you still aren't convinced, consult some textbook.
yeah, ignore me. I even deleted my own post! Got kinda confused.
 

CheesePlease

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So I thought.

Most of the perms/combs questions I've done are just rewordings of the same thing it's nice to see something different.

If it is asking what i think it's asking I'm a little stumped. But as the wording is so confusing already, I'll just make sure it's not 4^9 x 9! is it? I wouldn't really think so as that would be using repetitions.
 

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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.
 
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yea i get what you say lol. but seriously not worth the 3 marks... considerign the first part was worth 3 marks
 

Pwnage101

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yea i get what you say lol. but seriously not worth the 3 marks... considerign the first part was worth 3 marks
True - if you don't see it, it seems very daunting; but if you are exposed to this before and it comes up, the last line of my solution isn't too hard to see.

Can someone confirm with me, this topic can be in the HSC right?
It can and (almost always ) is.
 

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