permutations

[davieddy]

How many ways can you arrange 60 coloured tiles in a row,
15 different colours, 4 of each colour.

(Not sure whether this is very hard, but would
like to know the answer).

[davieddy]

60!/(4!)^15
= 1.648*10^61

I've temporarily forgotten the magic
"hide answer" instruction.

Not to worry.
D.

[mfgoode]

Originally Posted by davieddy

How many ways can you arrange 60 coloured tiles in a row,
15 different colours, 4 of each colour.

(Not sure whether this is very hard, but would
like to know the answer).

Do the rows have 4 different colours or does each row have the same coloured ties?

Mally

[davieddy]

Originally Posted by mfgoode

Do the rows have 4 different colours or does each row have the same coloured ties?

Mally

We have 4 tiles of each colour.
The "row" contains 60 spaces for tiles.
Can you explain why my spoiled answer gives
the number of distinct permutations?

[mfgoode]

Originally Posted by davieddy

We have 4 tiles of each colour.
The "row" contains 60 spaces for tiles.
Can you explain why my spoiled answer gives
the number of distinct permutations?

Off hand my intuition tells me your answer is erroneous, the very fact you are using an exponent (15) its not a common answer for Permutations or combinations. Exponents normally come into play with distributions unless I have read your problem wrong.

The common meaning would be 60 tiles arranged in 15 rows of 4 different colours each.
But you are calling it one 'row' of 60 spaces. That would more likely be a string of 60 spaces in one line

However if your answer is right I would certainly want to know how you get such a large figure?

Kindly clarify,

Mally

[grandpascorpion]

It's really a basic permutation problem.

Consider a row of 4 balls with two of each color: a and b.

There are 4!/(2!)^2 = 6 permutations.

a a b b
a b b a
b b a a
a b a b
b a b a
b a a b

The original question just involves larger values

[davieddy]

Originally Posted by mfgoode

However if your answer is right I would certainly want to know how you get such a large figure?

Mally

If all tiles had different colours, there would be 60! permutations.
If we take 4 of the colours and make them the same
(indistinguishable), then the 4! ways of rearranging these
tiles now count as one arrangement, reducing the number of
distinct arrangements to 60!/4!.
Do this for all 15 sets of 4 colours and we get 60!/4!^15.

It is the same principle as in deriving nCr = n!/(r!(n-r)!)

David

[m_f_h]

Originally Posted by mfgoode

... does each row ...

It was said : in a row.

[m_f_h]

Originally Posted by grandpascorpion

Consider a row of 4 balls with two of each color: a and b.
There are 4!/(2!)^2 = 6 permutations.

(hm. I checked it by counting all different permutations of 112233 - there are indeed 90=6!/2!^3.)

[wblipp]

The explanation I find most intuitive is to start by placing the four tiles of the first color. There are "60 choose 4" ways to do this, well known to be 60!/(56!*4!).

Next place the second color. For every combination of the first color there are "56 choose 4" choices. Then "52 choose 4" then "48 choose 4" etc. The numerator of each cancels the large factorial in the denominator of the previous, resulting in the simple expression.

[davieddy]

As may be deduced from the time I posted my
calculated answer, I had "intuited" the formula
almost before I had finished editing the question.
David