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# How many ways 6 rings can be worn in 4 fingers

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How many ways 6 rings can be worn in 4 fingers [#permalink]  19 Mar 2005, 17:30
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How many ways 6 rings can be worn in 4 fingers? Please explain.
VP
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each ring has a choice of 4 fingers, so no of ways.....4^6
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Re: PS: P&C problem [#permalink]  20 Mar 2005, 17:08
gmat2me2 wrote:
How many ways 6 rings can be worn in 4 fingers? Please explain.

Baner's answer assumes any finger can have any number of rings (which is perfect, since no restriction has been specified in the question)

Try the following interesting variations:

1) How many ways 6 rings can be worn in 4 fingers when a finger can have a maximum of 2 rings?
2) How many ways 6 rings can be worn in 4 fingers when a finger can have a maximum of 3 rings and each finger must have at least 1 ring?
3) How many ways 6 rings can be worn in 4 fingers when at least 3 fingers should have rings and any finger can't have more than 2 rings?

I don't know the actual answers, so would post the answers I calculate in a while.
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4^6 too

tough questions kapslock
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Re: PS: P&C problem [#permalink]  21 Mar 2005, 00:41
kapslock wrote:
gmat2me2 wrote:
How many ways 6 rings can be worn in 4 fingers? Please explain.

Baner's answer assumes any finger can have any number of rings (which is perfect, since no restriction has been specified in the question)

Try the following interesting variations:

1) How many ways 6 rings can be worn in 4 fingers when a finger can have a maximum of 2 rings?
2) How many ways 6 rings can be worn in 4 fingers when a finger can have a maximum of 3 rings and each finger must have at least 1 ring?
3) How many ways 6 rings can be worn in 4 fingers when at least 3 fingers should have rings and any finger can't have more than 2 rings?

I don't know the actual answers, so would post the answers I calculate in a while.

Alright Antmarvel, I am posting my calculations. Correct me if I am wrong (I just invented these problems on-the-fly) and post if you've got a better solution.

I am assuming the rings are all alike.

It would be interesting to solve for rings being distinct (Antmarvel - don't I just keep increasing the complexity? )

1) 8 places on 4 fingers, 6 rings, problem of choice of 6 from 8.
C(8, 6) = 28.

2) Since each finger must have one ring, lets place 1 ring in each finger. 4 rings gone, 2 remain, and now we have 4x2 = 8 positions (2 positions for each finger now remain - since each finger could accomodate a max of 3 rings). Again, a problem of selecting 2 out of 8, C(8, 2) = 28.

3) Case 1 : 3 fingers have rings. Place 1 ring each in these 3 fingers. You've 3 rings remaining now. Available positions = 3. Choices = C(3,2) = 3.

Case 2: All 4 fingers have rings.Place 1 ring in all 4 fingers. 2 rings remain, for 4 positions. Choices = C(4,2) = 6.

Thus total choices = 6 + 3 = 9.

I have some doubt on the solution for case 3. Antmarvel/anyone else would want to have a look please?

Thanks.
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This one is definitely too tough for me, I am still reviewing easy quant

However I hope HongHu, Ywelfried, banerjeea_98, rupstars and others to correct your post or to confirm the answers you've given because I want to know how to calculate this
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Re: PS: P&C problem [#permalink]  25 Mar 2005, 08:44
Wow so many rings and fingers.

I agree the answer to the original question is 4^6 if the rings are different from each other. It may be more complicated if the order of the rings on one finger is important. Would it be 4^6*P(6,6)?

If they are not different from each other, then we need to separate the 6 rings into four sections with three divisors. Total number of positions is 6+3=9 for 6 rings and 3 divisors. We need to pick three positions to place the three divisors. So the outcome would be C(9,3).

Now the different variations:

kapslock wrote:
1) How many ways 6 rings can be worn in 4 fingers when a finger can have a maximum of 2 rings?

Assuming they are not different from each other. There are 6 rings and 4 fingers so that means there could only be one finger without a ring if each finger can have a maximum of 2 rings. The outcome would be C(4,1) to decide which finger would be the bare finger, then the other three fingers would have 2 rings each.
Another type of outcome would be no bare fingers. In other words each finger would have at least 1 ring, then we need to pick two fingers to each have 1 more ring. The outcome would be C(4,2).
So my answer to this question is C(4,1)+C(4,2)

If they are different from each other, then order would become important too.
One bare finger: C(4,1)*P(6,6)
No bare fingers: C(4,2)*P(6,6)

Quote:
2) How many ways 6 rings can be worn in 4 fingers when a finger can have a maximum of 3 rings and each finger must have at least 1 ring?

Assuming they are not different. Each finger must have 1 ring so we only need to arrange the 2 extra ring. The total outcome would be C(5,3)=10. You could also do it this way: the two rings are in one finger: C(4,1). The two rings are in different fingers: C(4,2). Total outcomes=4+6=10.

If they are different, then we would need to take out the case where there is one or more bare fingers from the total outcome 4^6. This will also take care the maximum 3 condition.
4^6-3^6-2^6-1

Quote:
3) How many ways 6 rings can be worn in 4 fingers when at least 3 fingers should have rings and any finger can't have more than 2 rings?

Isn't it the same with question number 1)?
Re: PS: P&C problem   [#permalink] 25 Mar 2005, 08:44
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