marbles
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ISTER_REG
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For the first part (marbles of one color are indistinguishable):
We pick from each color one marble, so we have B (blue), W (white) , R (red), b (black). We can arrange these four different colored marbles (B,W,R,b) in 4! = 24 ways. Note: I assumed that we picked one marble from each color, this is (for me) not very clear from the task...
For the second part (marbles of one color are distinguishable):
Here we must then consider that each ball is now uniquely distinguishable by its color and a number. E.g. B1,R1,W1,b1 is then something different than B2,R1,W1,b1, (but the color is the same) where the letters stand for the colors and the number for the numbering of this marble. One must then further consider how many possibilities there are per color to draw a certain marble ...
We pick from each color one marble, so we have B (blue), W (white) , R (red), b (black). We can arrange these four different colored marbles (B,W,R,b) in 4! = 24 ways. Note: I assumed that we picked one marble from each color, this is (for me) not very clear from the task...
For the second part (marbles of one color are distinguishable):
Here we must then consider that each ball is now uniquely distinguishable by its color and a number. E.g. B1,R1,W1,b1 is then something different than B2,R1,W1,b1, (but the color is the same) where the letters stand for the colors and the number for the numbering of this marble. One must then further consider how many possibilities there are per color to draw a certain marble ...
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I wonder does this mean that any 4 in a row must contain the 4 colours?
(I changed the colours)!
[MATH] \underbrace{B \:R \: W\: G }\:B\: R\: W\: G\: B\: R\: W\: G\:...\\ B\:\underbrace{R\: W\: G \:B}\: R\: W\: G\: B\: R\: W\: G\:...\\ B\:R\:\underbrace{ W\: G \:B\: R}\: W\: G\: B\: R\: W\: G\:...\\ B\:R\: W\: \underbrace{G \:B\: R\: W}\: G\: B\: R\: W\: G\:...\\ B\:R\: W\: G \:\underbrace{B\: R\: W\: G}\: B\: R\: W\: G\:...\\[/MATH]
or does it mean that in a sequence of every fourth marble, no consecutive marbles have the same colour?
G G W R R W G B G G W W W B B B G G ...
or something else?
(I changed the colours)!
[MATH] \underbrace{B \:R \: W\: G }\:B\: R\: W\: G\: B\: R\: W\: G\:...\\ B\:\underbrace{R\: W\: G \:B}\: R\: W\: G\: B\: R\: W\: G\:...\\ B\:R\:\underbrace{ W\: G \:B\: R}\: W\: G\: B\: R\: W\: G\:...\\ B\:R\: W\: \underbrace{G \:B\: R\: W}\: G\: B\: R\: W\: G\:...\\ B\:R\: W\: G \:\underbrace{B\: R\: W\: G}\: B\: R\: W\: G\:...\\[/MATH]
or does it mean that in a sequence of every fourth marble, no consecutive marbles have the same colour?
G G W R R W G B G G W W W B B B G G ...
or something else?
ISTER_REG
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Ahh ok you have to translate, this explains in some cases the misunderstanding of those questions 
But, to my other question, do you have solutions, If so it is in some cases possible to come from the answer (only the number) to the calculation behind it. For example, if you have the answer to the second question here (as a number), I could help you to explain the calculation path
But, to my other question, do you have solutions, If so it is in some cases possible to come from the answer (only the number) to the calculation behind it. For example, if you have the answer to the second question here (as a number), I could help you to explain the calculation path
Ahh ok you have to translate, this explains in some cases the misunderstanding of those questions
But, to my other question, do you have solutions, If so it is in some cases possible to come from the answer (only the number) to the calculation behind it
The solution for the first case (a) is 4! (and i still do not completely understand why) and the solution for the second case (b) is 4!×(5!)^^4
ISTER_REG
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Assuming that I have understood the task correctly, the first part is about a permutation of 4 elements (The colours), the first colour has four possible placements, the second only three, the third two and the fourth only one (Imagine you would place the balls in four containers). This is then the relatively "typical" permutation, which then amounts to 4! = 4*3*2*1 = 24.
Now I understood the task, thank you!
ISTER_REG
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No problem, but are you sure that the solution for (b) is "4!×(5!)^^4" ? I would have rather thought of 4! * 5^4...
As my earlier post shows, for part 1, once you have determined the first four, the whole sequence is done. So there are 4! possibilities.
For question 2. The arrangements of the first four colours are 4! = 24 in number. The colours for the whole sequence are just a repeat of these 4. However because these balls are now all different, if you take the first group of four - for the first ball there are 5 choices of ball for that colour, for the second ball, there are 5 choices for a ball of that colour etc... so having decided the colour sequence, there are 5 * 5 * 5 *5 different combinations of ball for the first group of four. Similarly for the second group of four, there are now only 4 of each colour left, so there are 4 * 4 * 4 * 4 combinations. For the next group of four, 3 * 3 * 3 *3, the next group of four, 2 * 2 * 2 * 2 and the last group of four, there is no choice (you only have 4 balls left).
So: 4! is the number of different sequences of colour
[MATH]5^4 \times 4^4 \times 3^4 \times 2^4[/MATH] is the number of combinations of numbered ball for one particular sequence of colours:
In total: [MATH] 4! \times (5!)^4[/MATH]
For question 2. The arrangements of the first four colours are 4! = 24 in number. The colours for the whole sequence are just a repeat of these 4. However because these balls are now all different, if you take the first group of four - for the first ball there are 5 choices of ball for that colour, for the second ball, there are 5 choices for a ball of that colour etc... so having decided the colour sequence, there are 5 * 5 * 5 *5 different combinations of ball for the first group of four. Similarly for the second group of four, there are now only 4 of each colour left, so there are 4 * 4 * 4 * 4 combinations. For the next group of four, 3 * 3 * 3 *3, the next group of four, 2 * 2 * 2 * 2 and the last group of four, there is no choice (you only have 4 balls left).
So: 4! is the number of different sequences of colour
[MATH]5^4 \times 4^4 \times 3^4 \times 2^4[/MATH] is the number of combinations of numbered ball for one particular sequence of colours:
In total: [MATH] 4! \times (5!)^4[/MATH]
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This is my take on the questions, in more detail.
1. There are 4! permutations of four colours.
Eg. B R W G
since there are 4 choices for the colour of the first ball, then 3 for the second ball, 2 for the third ball, and the fourth is determined.
[MATH]4 \times 3 \times 2 \times 1[/MATH]
Now once you have a particular permutation for the colours of the first four: e.g. B R W G
since four consecutive balls must be different colours, the rest of the sequence must just repeat these colours:
[MATH]\boxed{\:B\:R \:W \:G\:}\boxed{\:B\:R \:W \:G\:}\boxed{\:B\:R \:W \:G\:}\boxed{\:B\:R \:W \:G\:}\boxed{\:B\:R \:W \:G\:}[/MATH]
since for [MATH]B\:\underbrace{R\: W\: G \:?\:}\:[/MATH] the next must be B again
[MATH]B\:R\:\underbrace{ W\: G \:B\: ? \:}\:[/MATH] the next must be R again etc…
[MATH]\therefore[/MATH] Answer to (1) is 4!
2. There are still 4! Arrangements of colours possible for the first four balls.
Once you have decided the sequence of colours for the first four
e.g. B R W G then you have determined the positions for the 5 B balls, the 5 R balls, the 5 W balls and the 5 G balls
e.g. for the sequence B R W G
Now, if each B ball is differently numbered, then there are 5! permutations of the B balls ( 5 choices for the first B ball in the list, then 4 choices for the second B ball in the list, then 3….)
The same applies for the R, the W, and the G balls.
[MATH]\therefore[/MATH] Total Number of arrangements:
number of arrangements of 4 colours[MATH] \: \times \: 5! \times 5! \times 5! \times 5![/MATH]
[MATH]=4!(5!)^4[/MATH]
1. There are 4! permutations of four colours.
Eg. B R W G
since there are 4 choices for the colour of the first ball, then 3 for the second ball, 2 for the third ball, and the fourth is determined.
[MATH]4 \times 3 \times 2 \times 1[/MATH]
Now once you have a particular permutation for the colours of the first four: e.g. B R W G
since four consecutive balls must be different colours, the rest of the sequence must just repeat these colours:
[MATH]\boxed{\:B\:R \:W \:G\:}\boxed{\:B\:R \:W \:G\:}\boxed{\:B\:R \:W \:G\:}\boxed{\:B\:R \:W \:G\:}\boxed{\:B\:R \:W \:G\:}[/MATH]
since for [MATH]B\:\underbrace{R\: W\: G \:?\:}\:[/MATH] the next must be B again
[MATH]B\:R\:\underbrace{ W\: G \:B\: ? \:}\:[/MATH] the next must be R again etc…
[MATH]\therefore[/MATH] Answer to (1) is 4!
2. There are still 4! Arrangements of colours possible for the first four balls.
Once you have decided the sequence of colours for the first four
e.g. B R W G then you have determined the positions for the 5 B balls, the 5 R balls, the 5 W balls and the 5 G balls
e.g. for the sequence B R W G
Now, if each B ball is differently numbered, then there are 5! permutations of the B balls ( 5 choices for the first B ball in the list, then 4 choices for the second B ball in the list, then 3….)
The same applies for the R, the W, and the G balls.
[MATH]\therefore[/MATH] Total Number of arrangements:
number of arrangements of 4 colours[MATH] \: \times \: 5! \times 5! \times 5! \times 5![/MATH]
[MATH]=4!(5!)^4[/MATH]