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In how many ways can 4 identical red balls and 5 different

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In how many ways can 4 identical red balls and 5 different [#permalink] New post 05 Oct 2005, 14:47
In how many ways can 4 identical red balls and 5 different green balls be arranged in a line such that no two red balls are together?
5*5!
5*4!*5!
6*4!*5!
6*5!
2*9!
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 [#permalink] New post 05 Oct 2005, 17:11
B.

With no restrictions 4R(identical) and 5G (different) can be arranged in:
(4 + 5) ! / 4! or 9!/4!

the number of arrangments where all red balls are together are:
5! (for all Green) * 6 {once all green balls are arranged, all red together can be put at 6 different positions}

therefore, all arrangments where no two red balls together =
9!/4! - 6 * 5! = 14400 = 5*4!*5!
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 [#permalink] New post 06 Oct 2005, 03:35
Your reasoning is correct but I think you misunderstood the question. "No two red balls are together" is not the same as "the four red balls are not all together".

Anyway, I do not get any of the answer choices:

There are 5! different ways to arrange the green balls.
For each of these arrangements there are C(6,2) possible arrangements of red balls so that no red balls are together.

Therefore 15*5!

Does anyone have a different result?
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 [#permalink] New post 06 Oct 2005, 06:30
I would say A.

You've got 5! ways the green balls can be arranged by themselves, since the reds are identical their order does not matter. There are 5 ways to arrange the greens so that no two reds are together therefore 5x5! should give you the right number of combos.

GGRGRGRGR = 5! combos
GRGGRGRGR = 5!
GRGRGGRGR = 5!
GRGRGRGGR = 5!
GRGRGRGRG = 5!
____
5X5! total combos

B.
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 [#permalink] New post 06 Oct 2005, 07:28
OA is 6*5!

Anyone knows a simple methos to arrive at that?
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 [#permalink] New post 06 Oct 2005, 09:39
rchadha wrote:
OA is 6*5!

Anyone knows a simple methos to arrive at that?


Actually, its fairly simple ->
Identical red balls can be arranged in only 1 set way. That is a gap between each of them.
The 5 different green balls amongst themselves can be arranged in 5! ways and 6 ways between the gaps created by Red as shown by bsmith75 which is missing one combo (see bold).

GGRGRGRGR = 5! combos
GRGGRGRGR = 5!
GRGRGGRGR = 5!
GRGRGRGGR = 5!
GRGRGRGRG = 5!
RGRGRGRGG

I.e 6*5!*1 = 6*5!
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 [#permalink] New post 06 Oct 2005, 10:22
vikramm wrote:
rchadha wrote:
OA is 6*5!

Anyone knows a simple methos to arrive at that?


Actually, its fairly simple ->
Identical red balls can be arranged in only 1 set way. That is a gap between each of them.
The 5 different green balls amongst themselves can be arranged in 5! ways and 6 ways between the gaps created by Red as shown by bsmith75 which is missing one combo (see bold).

GGRGRGRGR = 5! combos
GRGGRGRGR = 5!
GRGRGGRGR = 5!
GRGRGRGGR = 5!
GRGRGRGRG = 5!
RGRGRGRGG

I.e 6*5!*1 = 6*5!

How did you get 6 ways to put 4 identical Red balls.

We have 6 positions available:
X G X G X G X G X G X

in these 6 position we need to put 4 Red balls (order not important)
wouldn't it be: 6C4 = 15

for example, why did you exclude these arrangements:
RGRGRGGRG
RGRGRGGGR
RGRGGRGRG

Last edited by duttsit on 06 Oct 2005, 11:12, edited 1 time in total.
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 [#permalink] New post 06 Oct 2005, 11:03
duttsit wrote:
vikramm wrote:
rchadha wrote:
OA is 6*5!

Anyone knows a simple methos to arrive at that?


Actually, its fairly simple ->
Identical red balls can be arranged in only 1 set way. That is a gap between each of them.
The 5 different green balls amongst themselves can be arranged in 5! ways and 6 ways between the gaps created by Red as shown by bsmith75 which is missing one combo (see bold).

GGRGRGRGR = 5! combos
GRGGRGRGR = 5!
GRGRGGRGR = 5!
GRGRGRGGR = 5!
GRGRGRGRG = 5!
RGRGRGRGG

I.e 6*5!*1 = 6*5!

How did you get 6 ways to put 4 identical Red balls.

We have 6 positions available:
X G X G X G X G X G X

in these 6 position we need to put 4 Red balls (order not important)
wouldn't it be: 6C4 = 10

for example, why did you exclude these arrangements:
RGRGRGGRG
RGRGRGGGR
RGRGGRGRG


Thanks duttsit, You are correct... my bad...
The combinations you list are valid...
Trying to avoid matching the answers... but just extending your method...
Why do we need 6C4? The four balls are identical.. shouldn't it be 6C1?
If we remove the Green Balls in the mix scenario... 4 identical Red balls together can be arranged in only one possible way.. as they are identical. Right? So, going back to -> X G X G X G X G X G X -> It doesnt matter which Red Ball goes where (as long as it is not next to each other) which means 6C1 ways... am I missing anything?
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 [#permalink] New post 06 Oct 2005, 11:07
There are actually 15 arrangements:

GGRGRGRGR
GRGGRGRGR
GRGRGGRGR
GRGRGRGGR
GRGRGRGRG
RGGGRGRGR
RGGRGGRGR
RGGRGRGGR
RGGRGRGRG
RGRGGGRGR
RGRGGRGGR
RGRGGRGRG
RGRGRGGGR
RGRGRGGRG
RGRGRGRGG

C(6,2)=15 Combinations with repetition of five elements taken two at a time.
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 [#permalink] New post 06 Oct 2005, 11:19
jdtomatito wrote:
There are actually 15 arrangements:

GGRGRGRGR
GRGGRGRGR
GRGRGGRGR
GRGRGRGGR
GRGRGRGRG
RGGGRGRGR
RGGRGGRGR
RGGRGRGGR
RGGRGRGRG
RGRGGGRGR
RGRGGRGGR
RGRGGRGRG
RGRGRGGGR
RGRGRGGRG
RGRGRGRGG

C(6,2)=15 Combinations with repetition of five elements taken two at a time.


Green Balls are not identical... you have many more combinations if you considered G as G1, G2, G3, G4, G5...
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 [#permalink] New post 07 Oct 2005, 04:18
As I said before, the solution is C(6,2)*5!

You have considered all the possibilities for the green balls in the 5!
  [#permalink] 07 Oct 2005, 04:18
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