20070707, 19:57  #1 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
permutations
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). Last fiddled with by davieddy on 20070707 at 19:59 
20070707, 20:07  #2 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
60!/(4!)^15
= 1.648*10^61 I've temporarily forgotten the magic "hide answer" instruction. Not to worry. D. Last fiddled with by davieddy on 20070707 at 20:32 Reason: But no more:) 
20070712, 09:30  #3 
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
Coloured rows

20070712, 11:54  #4 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 

20070712, 16:03  #5  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Rows.
Quote:
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 

20070712, 16:12  #6 
Jan 2005
Transdniestr
503_{10} Posts 
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 
20070712, 17:08  #7  
"Lucan"
Dec 2006
England
14512_{8} Posts 
Quote:
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!(nr)!) David 

20070713, 00:58  #8 
Feb 2007
2^{4}×3^{3} Posts 

20070713, 01:03  #9 
Feb 2007
432_{10} Posts 

20070713, 03:51  #10 
"William"
May 2003
New Haven
4500_{8} Posts 
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. 
20070713, 05:22  #11 
"Lucan"
Dec 2006
England
194A_{16} Posts 
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 
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