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If I have four balls; 2 blue, 1 red and 1 green, how many possible ways do I have to arrange them, without repetition of a colour sequence?

Help pls

Help pls

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#2

(Original post by

If I have four balls; 2 blue, 1 red and 1 green, how many possible ways do I have to arrange them, without repetition of a colour sequence?

Help pls

**Kalabamboo**)If I have four balls; 2 blue, 1 red and 1 green, how many possible ways do I have to arrange them, without repetition of a colour sequence?

Help pls

BBRG is one

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So..I am trying to about it via combinations/using factorial etc but am not sure how to approach it.

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#5

(Original post by

That's the long way of doing it.

So..I am trying to about it via combinations/using factorial etc but am not sure how to approach it.

**Kalabamboo**)That's the long way of doing it.

So..I am trying to about it via combinations/using factorial etc but am not sure how to approach it.

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(Original post by

18 ( I think......not sure)

**joyce365**)18 ( I think......not sure)

Not sure what they did though - I am sure there is a quicker way to do this question without listing the whole lot out.

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#8

Assume you have 4 unique objects, these can be ordered in 4! ways (=24) BUT two of the unique objects are interchangeable, hence every sequence will actually appear twice, to account for this divide the total number of possible arrangements by 2. = 12 tada~

EX: B1 B2 R G and B2 B1 R G are 'different' but since B1 and B2 are both blue we only have 1 sequence.

EX: B1 B2 R G and B2 B1 R G are 'different' but since B1 and B2 are both blue we only have 1 sequence.

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(Original post by

Assume you have 4 unique objects, these can be ordered in 4! ways (=24) BUT two of the unique objects are interchangeable, hence every sequence will actually appear twice, to account for this divide the total number of possible arrangements by 2. = 12 tada~

EX: B1 B2 R G and B2 B1 R G are 'different' but since B1 and B2 are both blue we only have 1 sequence.

**Qkzr**)Assume you have 4 unique objects, these can be ordered in 4! ways (=24) BUT two of the unique objects are interchangeable, hence every sequence will actually appear twice, to account for this divide the total number of possible arrangements by 2. = 12 tada~

EX: B1 B2 R G and B2 B1 R G are 'different' but since B1 and B2 are both blue we only have 1 sequence.

just trying to learn as many ways to do this question

is it possible to do this question in terms of factorial and notation in form NCR or NPR stuff like that? lemme know your thoughts please

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#10

(Original post by

Hey thanks a lot this makes sense

just trying to learn as many ways to do this question

is it possible to do this question in terms of factorial and notation in form NCR or NPR stuff like that? lemme know your thoughts please

**Kalabamboo**)Hey thanks a lot this makes sense

just trying to learn as many ways to do this question

is it possible to do this question in terms of factorial and notation in form NCR or NPR stuff like that? lemme know your thoughts please

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(Original post by

You can also think of it as choosing the places for the blue balls and then arranging the red and the green balls. We can choose the places for the two blue balls in ways then we can arrange the remaining balls in ways. So there are 12 ways of doing this by the multiplication principle.

**Cryptokyo**)You can also think of it as choosing the places for the blue balls and then arranging the red and the green balls. We can choose the places for the two blue balls in ways then we can arrange the remaining balls in ways. So there are 12 ways of doing this by the multiplication principle.

When they say "without repetition of a colour sequence" what do they mean in this context

Also, I didn't quite understand your reply back

So just to let you know at what level I am at in terms of my understanding - I've just explained my understanding of the question below

Way 1: just do 10C6

Way 2: 10 x 9 x 8 x7 x 6 x5 = 151200 permutations

6 x 5 x 4 x 3 x 2 x1 = 720 (not rlly sure what this stands for but can be used to work out combinations)

151200/720 = 210 combinations

I trying to apply the same thing to the question in this thread but not rlly understanding it

Could you pls kindly help?

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#12

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#13

(Original post by

Hi thanks a lot for replying - this is exactly what I meant but i struggle a bit with this

When they say "without repetition of a colour sequence" what do they mean in this context

Also, I didn't quite understand your reply back

So just to let you know at what level I am at in terms of my understanding - I've just explained my understanding of the question below

Way 1: just do 10C6

Way 2: 10 x 9 x 8 x7 x 6 x5 = 151200 permutations

6 x 5 x 4 x 3 x 2 x1 = 720 (not rlly sure what this stands for but can be used to work out combinations)

151200/720 = 210 combinations

I trying to apply the same thing to the question in this thread but not rlly understanding it

Could you pls kindly help?

**Kalabamboo**)Hi thanks a lot for replying - this is exactly what I meant but i struggle a bit with this

When they say "without repetition of a colour sequence" what do they mean in this context

Also, I didn't quite understand your reply back

So just to let you know at what level I am at in terms of my understanding - I've just explained my understanding of the question below

Way 1: just do 10C6

Way 2: 10 x 9 x 8 x7 x 6 x5 = 151200 permutations

6 x 5 x 4 x 3 x 2 x1 = 720 (not rlly sure what this stands for but can be used to work out combinations)

151200/720 = 210 combinations

I trying to apply the same thing to the question in this thread but not rlly understanding it

Could you pls kindly help?

- Let us consider the example you gave first and come back to the other question in a moment. When you are picking a team of players the players are all distinct but the order in which they appear in the team does not matter. We simply need to CHOOSE 6 players from a set of 10. And this is the giveaway for the way you first solved it (way 1).
- But this approach cannot be used for our original problem but first let us clarify what a colour sequence is. I understand a colour sequence to be the order in which the colours appear i.e. BBGR, BGRB, GBRB, .. etc. If we were given the colour sequence BGBR and swapped the positions of the blue balls then we would still have BGBR which is the same colour sequence (i.e a repetition) but if we swap the green and the red we have BRBG which is a different colour sequence. Is this clear?
- Now let us see how the first way (using binomial cofficients) of attempting this problem fails. In the problem we want to find the number of ways we can ORDER the colours and the word order is key here. When we use the binomial coefficient the order does not matter and also it does not work (in general!! [although there are other ways]) for dealing with identical objects (the two blue balls in this example). So quite easily you get stuck.
- Now your best bet from what you know to answer this problem is using a slightly adapted version of the 2nd way you answered your example. If we had 4 distinct balls we could order them 4x3x2x1 = 24 ways. Agree? But 2 of them are blue and as we said before swapping the positions of the blue balls gave the same colour sequence. So for each of our arrangements we have just found we have another different arrangement with an equivalent colour sequence. So we must divide the answer by 2 to get rid of this symmetry. Hence there are 12 possible ways.
- Hope this helps! Combinatorics is quite annoying to get your head around! The solution I gave earlier used a few extra little tricks which are handy on more tricky problems.

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#14

You have 10 unique objects (players) and you want the total number of possible combinations (combinations are groups are independent of order). The correct thing to do is 10 C 6.

The key point here is that a combination is a selection and it is independent of order. This means that the team L M N O P Q and Q L M N O P are still the same. This is obvious with teams as regardless of the order you pick the players in you still have the same players on your team.

It is possible to count the number of possible 6 player combinations starting with permutations.

You can create 10P6 = 10 x 9 x 8 x 7 x 6 x 5 unique 6 player sequences (permutations). Since there are 6 objects in our permutation the same 6 player combination of just for example L M N O P Q R will appear 6! times. To account for this you do 10 P 6 / 6!.

Notice that 10P6/6! is the same as 10C6. nPr/r! = nCr

The key point here is that a combination is a selection and it is independent of order. This means that the team L M N O P Q and Q L M N O P are still the same. This is obvious with teams as regardless of the order you pick the players in you still have the same players on your team.

It is possible to count the number of possible 6 player combinations starting with permutations.

You can create 10P6 = 10 x 9 x 8 x 7 x 6 x 5 unique 6 player sequences (permutations). Since there are 6 objects in our permutation the same 6 player combination of just for example L M N O P Q R will appear 6! times. To account for this you do 10 P 6 / 6!.

Notice that 10P6/6! is the same as 10C6. nPr/r! = nCr

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(Original post by

**Cryptokyo**)- Let us consider the example you gave first and come back to the other question in a moment. When you are picking a team of players the players are all distinct but the order in which they appear in the team does not matter. We simply need to CHOOSE 6 players from a set of 10. And this is the giveaway for the way you first solved it (way 1).
- But this approach cannot be used for our original problem but first let us clarify what a colour sequence is. I understand a colour sequence to be the order in which the colours appear i.e. BBGR, BGRB, GBRB, .. etc. If we were given the colour sequence BGBR and swapped the positions of the blue balls then we would still have BGBR which is the same colour sequence (i.e a repetition) but if we swap the green and the red we have BRBG which is a different colour sequence. Is this clear?
- Now let us see how the first way (using binomial cofficients) of attempting this problem fails. In the problem we want to find the number of ways we can ORDER the colours and the word order is key here. When we use the binomial coefficient the order does not matter and also it does not work (in general!! [although there are other ways]) for dealing with identical objects (the two blue balls in this example). So quite easily you get stuck.
- Now your best bet from what you know to answer this problem is using a slightly adapted version of the 2nd way you answered your example. If we had 4 distinct balls we could order them 4x3x2x1 = 24 ways. Agree? But 2 of them are blue and as we said before swapping the positions of the blue balls gave the same colour sequence. So for each of our arrangements we have just found we have another different arrangement with an equivalent colour sequence. So we must divide the answer by 2 to get rid of this symmetry. Hence there are 12 possible ways.
- Hope this helps! Combinatorics is quite annoying to get your head around! The solution I gave earlier used a few extra little tricks which are handy on more tricky problems.

this makes much sense! With the solution you gave earlier , you know when you said "You can also think of it as choosing the places for the blue balls and then arranging the red and the green balls. We can choose the places for the two blue balls in ways

**then we can arrange the remaining balls in**

**ways.**So there are 12 ways of doing this by the multiplication principle."

Ik im silly for asking this but I don't understand how there are 2 ways of doing this and what 2! has to do with this

also why don't we involve permutations in this?

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#16

(Original post by

Ahh thanks a lot -you cleared so many thing i didn't even know if i was confused about for me lol

this makes much sense! With the solution you gave earlier , you know when you said "You can also think of it as choosing the places for the blue balls and then arranging the red and the green balls. We can choose the places for the two blue balls in ways

Ik im silly for asking this but I don't understand how there are 2 ways of doing this and what 2! has to do with this

also why don't we involve permutations in this?

**Kalabamboo**)Ahh thanks a lot -you cleared so many thing i didn't even know if i was confused about for me lol

this makes much sense! With the solution you gave earlier , you know when you said "You can also think of it as choosing the places for the blue balls and then arranging the red and the green balls. We can choose the places for the two blue balls in ways

**then we can arrange the remaining balls in****ways.**So there are 12 ways of doing this by the multiplication principle."Ik im silly for asking this but I don't understand how there are 2 ways of doing this and what 2! has to do with this

also why don't we involve permutations in this?

Nah nothing is a silly question , this stuff is difficult to get your head round - it certainly took me a while! This is indeed permutations but with repetitions so there is an extra layer of subtlety here. I have solved it without permutations as such - a lot of problems can be done in many ways!

I'll explain my way as we have sort of unofficially covered the permutations with repetitions argument when we found the number of permutations of the 4 balls then accounted for the symmetry caused by the ability to maintain the colour sequence by switching the two balls.

The way my method works is as follows:

- We say we have four positions to fill: _ _ _ _
- Then we CHOOSE the two positions from the four positions in which the blue balls will go. There are 4 choose 2 ways of doing this - which is 6 ways.
- Once we have chosen the places for the blue balls we put them there. Now we have two spaces remaining and the scenario we are in looks something like: B _ B _, or _ B B _ or B _ _ B etc.
- Now we just need to choose the order for the red and the green balls. Starting from the leftmost most empty position, we have 2 choices for the ball (red or green). Once we have decided this, the other position must be filled with the other ball. So there are 2x1 ways of doing this. I write this as 2! (that is where it comes from).
- Then we multiply the two numbers to give 6x2=12 arrangements.

So the key point here is sometimes it is easier to choose the positions in an arrangement for a set of objects instead of trying to immediately decide the order in which they should go. Here the order of the blue balls does not matter so this technique works rather nicely. There are some more nasty examples where this is used but the set of objects which are non-identical and a few other tricks are used.

An example of such a problem would be: Find the number of arrangements of the word CABBAGE such that no two vowels are adjacent to each other (e.g. CABBEGA is fine but CBAEBAG is not).

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(Original post by

Glad it helped!

Nah nothing is a silly question , this stuff is difficult to get your head round - it certainly took me a while! This is indeed permutations but with repetitions so there is an extra layer of subtlety here. I have solved it without permutations as such - a lot of problems can be done in many ways!

I'll explain my way as we have sort of unofficially covered the permutations with repetitions argument when we found the number of permutations of the 4 balls then accounted for the symmetry caused by the ability to maintain the colour sequence by switching the two balls.

The way my method works is as follows:

So the key point here is sometimes it is easier to choose the positions in an arrangement for a set of objects instead of trying to immediately decide the order in which they should go. Here the order of the blue balls does not matter so this technique works rather nicely. There are some more nasty examples where this is used but the set of objects which are non-identical and a few other tricks are used.

An example of such a problem would be: Find the number of arrangements of the word CABBAGE such that no two vowels are adjacent to each other (e.g. CABBEGA is fine but CBAEBAG is not).

**Cryptokyo**)Glad it helped!

Nah nothing is a silly question , this stuff is difficult to get your head round - it certainly took me a while! This is indeed permutations but with repetitions so there is an extra layer of subtlety here. I have solved it without permutations as such - a lot of problems can be done in many ways!

I'll explain my way as we have sort of unofficially covered the permutations with repetitions argument when we found the number of permutations of the 4 balls then accounted for the symmetry caused by the ability to maintain the colour sequence by switching the two balls.

The way my method works is as follows:

- We say we have four positions to fill: _ _ _ _
- Then we CHOOSE the two positions from the four positions in which the blue balls will go. There are 4 choose 2 ways of doing this - which is 6 ways.
- Once we have chosen the places for the blue balls we put them there. Now we have two spaces remaining and the scenario we are in looks something like: B _ B _, or _ B B _ or B _ _ B etc.
- Now we just need to choose the order for the red and the green balls. Starting from the leftmost most empty position, we have 2 choices for the ball (red or green). Once we have decided this, the other position must be filled with the other ball. So there are 2x1 ways of doing this. I write this as 2! (that is where it comes from).
- Then we multiply the two numbers to give 6x2=12 arrangements.

So the key point here is sometimes it is easier to choose the positions in an arrangement for a set of objects instead of trying to immediately decide the order in which they should go. Here the order of the blue balls does not matter so this technique works rather nicely. There are some more nasty examples where this is used but the set of objects which are non-identical and a few other tricks are used.

An example of such a problem would be: Find the number of arrangements of the word CABBAGE such that no two vowels are adjacent to each other (e.g. CABBEGA is fine but CBAEBAG is not).

you know where you say "This is indeed permutations but with repetitions so there is an extra layer of subtlety here. I have solved it without permutations as such - a lot of problems can be done in many ways!"

and where you say "You can also think of it as choosing the places for the blue balls and then arranging the red and the green balls. We can choose the places for the two blue balls in ways

**then we can arrange the remaining balls in**

**ways.**So there are 12 ways of doing this by the multiplication principle."

so we are clearly doing 4C2 x 2C1 x 1C1 to get 12 but what I don't understand is that since you said it is permutations, why don't we do 4P2 x 2P1 x 1P1?

Just don't get the difference between 4C2 x 2C1 x 1C1 and 4P2 x 2P1 x 1P1? Like why do we do the first one in this question?

Thanks a lot

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#18

(Original post by

Ah thank you very much - really appreciate it

you know where you say "This is indeed permutations but with repetitions so there is an extra layer of subtlety here. I have solved it without permutations as such - a lot of problems can be done in many ways!"

and where you say "You can also think of it as choosing the places for the blue balls and then arranging the red and the green balls. We can choose the places for the two blue balls in ways

so we are clearly doing 4C2 x 2C1 x 1C1 to get 12 but what I don't understand is that since you said it is permutations, why don't we do 4P2 x 2P1 x 1P1?

Just don't get the difference between 4C2 x 2C1 x 1C1 and 4P2 x 2P1 x 1P1? Like why do we do the first one in this question?

Thanks a lot

**Kalabamboo**)Ah thank you very much - really appreciate it

you know where you say "This is indeed permutations but with repetitions so there is an extra layer of subtlety here. I have solved it without permutations as such - a lot of problems can be done in many ways!"

and where you say "You can also think of it as choosing the places for the blue balls and then arranging the red and the green balls. We can choose the places for the two blue balls in ways

**then we can arrange the remaining balls in****ways.**So there are 12 ways of doing this by the multiplication principle."so we are clearly doing 4C2 x 2C1 x 1C1 to get 12 but what I don't understand is that since you said it is permutations, why don't we do 4P2 x 2P1 x 1P1?

Just don't get the difference between 4C2 x 2C1 x 1C1 and 4P2 x 2P1 x 1P1? Like why do we do the first one in this question?

Thanks a lot

^^ Potentially this is easier to understand as a method rather than my ramblings.

My argument was subtly different. I am saying. I CHOOSE 2 out of the 4 places for the blue balls. Then I find the number of permutations for the red and the green ball in which they can placed in the two remaining places - which is 2P2 = 2 = 2! ways. So my solution is more strictly 4C2 x 2P2 = 12.

We can't do 4P2 as the permutation formula on works for distinguishable objects but here the blue balls are indistinguishable. So 4P2 ends up double counting all the permutations with a blue ball in them. And then after this we don't know if we have placed a blue ball or not. So the next part of our logic may very well break down. This is ultimately why we do the first one as the logic is unclear.

However, if you wished to use permutations for this problem the logic would again be different and the multiplication would look again different.

The main point is which of the permutations and combinations give you the most logical and straightforward approach to the problem. Both can be used to solve most of these problems they just require the correct manipulation of the formulae. And knowing which one is more straightforward to use mainly comes from experience and you'll get into a certain way of looking at the problems.

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#19

(Original post by

Hey I've seen you somewhere before. Aren't you the guy that doesn't like Oxbridge or something?

**Anonymouspsych**)Hey I've seen you somewhere before. Aren't you the guy that doesn't like Oxbridge or something?

I am not anti-Oxbridge but it is the wrong place for some degrees and some people.

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#20

(Original post by

nope it is 12

Not sure what they did though - I am sure there is a quicker way to do this question without listing the whole lot out.

**Kalabamboo**)nope it is 12

Not sure what they did though - I am sure there is a quicker way to do this question without listing the whole lot out.

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