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).

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

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

Not to worry.
D.

Coloured rows
 

[QUOTE=davieddy;109822]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).[/QUOTE]

:confused:

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

Mally :coffee:

[quote=mfgoode;110142]:confused:

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

Mally :coffee:[/quote]

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?

Rows.
 

[QUOTE=davieddy;110148]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?[/QUOTE]

:smile:

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 :coffee:

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

[quote=mfgoode;110185]:smile:

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

Mally :coffee:[/quote]
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

[quote=mfgoode;110142]... does each row ...[/quote]
It was said : in [U][B]a[/B][/U] row.

[quote=grandpascorpion;110189]Consider a row of 4 balls with two of each color: a and b.
There are 4!/(2!)^2 = 6 permutations.[/quote]
(hm. I checked it by counting all different permutations of 112233 - there are indeed 90=6!/2!^3.)

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.
:wblipp:

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

[QUOTE=davieddy;110196]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[/QUOTE]

:whistle:

Thank you Davie for the elucidation.
I am more familiar with the nCr notation hence my confusion with the exponent.

Mally :coffee:

partitions
 

if we have p red, q green and r blue tiles, there are
(p+q+r)!/(p!q!r!) arrangements.
nCr can be seen as a special case of this formula.
(As can my formula by taking p=q=r)

David

Equivalence!
 

[QUOTE=davieddy;110196]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[/QUOTE]

:rolleyes:

Kindly bear with me Davie. I gave up teaching HSC (equivalent to A level maths) as far back as 1999 during my first two years of retirement. So Im rusty like your blade is blunt.

I understand that both (60!/4!)^15 and nCr are equivalent.

Thus lets put it as (60!/4!)^14 * 60!/4! =nCr = n!/4!/(n - 4)!

n!/4! (60!/4!) is common to both sides so cancel it out

We then get (60!/4!)^14 = 1/(n-4)!

We thus get a gigantic number on the L.H.S. > 1 = R.H.S. < 1 ?

Where is the mistake Sir ?

Mally :coffee:

[QUOTE=mfgoode;110290]:
I understand that both (60!/4!)^15
Where is the mistake Sir ?[/QUOTE]
[QUOTE= davieddy]we get 60!/4!^15.[/QUOTE]

Mally,

60!/4!^15 is not the same expression as (60!/4!)^15

[quote=mfgoode;110290]
I understand that both (60!/4!)^15 and nCr are equivalent.

Thus lets put it as (60!/4!)^14 * 60!/4! =nCr = n!/4!/(n - 4)!
...
Where is the mistake Sir ?

Mally :coffee:[/quote]

They are not equal to each other whether
you interpret 60!/(4!^15) correctly or not.
I wouldn't say "equivalent" either, but see
my "partitions" post.

David

My error.
 

[QUOTE=Wacky;110291]Mally,

60!/4!^15 is not the same expression as (60!/4!)^15[/QUOTE]

:surprised

Thank you Wacky for pointing that out. Now its a horse with a different colour! Just out of curiousity I computed the number. It has 11 trailing zeroes!

Mally :coffee:

Mally,
You placed the brackets in (60!/4!)^15.
To interpret 60!/4!^15 as the intended 60!/(4!^15)
requires an understanding of "prioirity" in expressions.
You may have heard of "BODMAS" regarding brackets,
division/multiplication and addition/subtraction.
Well, "BEDMAS" is a better mnenonic, where E stands
for exponentiation instead of nothing in particular.

David

PS Your mistake is tantamount to confusing
2+3*4(=14) with (2+3)*4(=20)

flogging a (differently coloured) horse
 

[QUOTE=mfgoode;110402]: Just out of curiousity I computed the number. It has 11 trailing zeroes!

Mally :coffee:[/QUOTE]

so did I. Of course many people here have access to arbitrary precision calculators making this sort of calculation trivial, but i don't so it took me a while, and i resorted to excel in the end. How did you do it?

Richard

[quote=mfgoode;110402]:surprised

Thank you Wacky for pointing that out. Now its a horse with a different colour! Just out of curiousity I computed the number. It has 11 trailing zeroes!

Mally :coffee:[/quote]

What is the colour of the horse to which you are referring?

David

Bedmas.
 

[QUOTE=davieddy;110414]Mally,
You placed the brackets in (60!/4!)^15.
To interpret 60!/4!^15 as the intended 60!/(4!^15)
requires an understanding of "prioirity" in expressions.
You may have heard of "BODMAS" regarding brackets,
division/multiplication and addition/subtraction.
Well, "BEDMAS" is a better mnenonic, where E stands
for exponentiation instead of nothing in particular.

David

PS Your mistake is tantamount to confusing
2+3*4(=14) with (2+3)*4(=20)[/QUOTE]

:smile:

Thank you Davie, thats a new one for me.

Easy to remember too -- Bed Mistress as Sweetheart! :missingteeth:

Mally :coffee:

[QUOTE=Richard Cameron;110421]so did I. Of course many people here have access to arbitrary precision calculators making this sort of calculation trivial, but i don't so it took me a while, and i resorted to excel in the end. How did you do it?

Richard[/QUOTE]:

smile:

Will be glad to oblige Richard!

I use 'factoris'. You can google it easily and click on it. It's one of the early lessons among many others I learnt from this forum. It can crack out any number and its factors and this particular number has many factors and it even certifies if they are prime.

I dont go in for the fancy stuff like Pari etc

Good luck to your computing!

Mally :coffee:

[quote=mfgoode;110429]:smile:

Thank you Davie, thats a new one for me.

Easy to remember too -- Bed Mistress as Sweetheart! :missingteeth:

Mally :coffee:[/quote]

I invented it on the spur of the moment.
Guessed you'd like the "BED" part of it tho:lol:

David

Horses!
 

[QUOTE=davieddy;110422]What is the colour of the horse to which you are referring?

David[/QUOTE]

:smile:

It's white and has a story that you will be mystified by!

Back in the old days -50's - Mumbai (then Bombay) was full of 'Victorias' a cheap form of transport. You know a horse driven carriage with a unique styling- thanks to British ingenuity.

Well I was dating a girl with animal passions and she always wanted to ride in a carriage drawn by a white horse. Now white horses here were rare, being reserved for the aristocrat's stables and meant for circuses, racing and the rest. So we were indeed lucky to get one on the roads.

Years later I realised when I read an account on Czarina Katherina of Russia why she was so mystified by white horses!

About the same time Simone du Beavoir ? came out with her classic
on "The second sex' ? (I hope I have got that right.) Yeah and she explained it very well why little girls are mystified by little boys piddling standing up as they cant do it. Do you get the connection ? :wink: :whistle:

Mally :coffee:

Talking of changing horses in midstream
Mally, I sometimes forget what thread
we are on about;-)

David

:grin:

Sorry Davie and to the others but its the same with me.

When someone put a cross thread reference I was at a loss as to which thread it is in.

Well I'm calling it a day as I have a date with Argentina at 02.30 this morning and before Wacky shunts us off by closing this thread I better wamoose!

Mally :coffee:

[quote=mfgoode;110436]:grin:

Sorry Davie and to the others but its the same with me.

When someone put a cross thread reference I was at a loss as to which thread it is in.

Well I'm calling it a day as I have a date with Argentina at 02.30 this morning and before Wacky shunts us off by closing this thread I better wamoose!

Mally :coffee:[/quote]

I hope Argentina is good in bed:lol:

I thought she was a country (silly me)

David

[quote]...many people here have access to arbitrary precision calculators making this sort of calculation trivial, but i don't...[/quote]You can get "bc" for many different operating systems, including Windows, for free.

[URL]http://www.gnu.org/software/bc/[/URL]
[URL]http://gnuwin32.sourceforge.net/packages/bc.htm[/URL]

[code]$ bc
bc 1.06
Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.

define f(n) {if (n<=1) return (1); return (n*f(n-1))}

f(60)/f(4)^15
16481860109099054170027351546863185027378649716962500000000000

(f(60)/f(4))^15
12573440042967022554399646208738393172034538801875083669851005228747\
34260499677802720217276602439941962952775943942192052494930241655552\
62988852813160436703937901438163944159657618605216019694443174340910\
80457076447665707640398537934869440661181972976087953472785671437878\
71015943867854569107314256518536445125099952602449526254621897013471\
32032091711111832207259701768251616201127088495304030289490118189969\
37410974856698490766750942513641548687909045876044741858114128298525\
63014256430817573572400861631986537315373773779611014601613776922143\
06238315540810543138847144796370924882712878726422058701161408383263\
01835566252031881485329496457221387236297759114234540997798322156682\
84602634533981344000247085914741074144957406661117488023730682213850\
07838520468435642421480728494285441701884814371112389552635221933768\
58717440961052368825456019332050556539071821895499766721339652921559\
51182724296526639731668086402966517483252992314732224346257399332575\
41901181007987099285456439721705534118715981824000000000000000000000\
00000000000000000000000000000000000000000000000000000000000000000000\
00000000000000000000000000000000000000000000000000000000000000000000\
00000000000000000000000000000000000000000000000000000[/code][SIZE=1]Here is a funny link that is vaguely related to this thread:

[URL="http://www.willamette.edu/%7Efruehr/haskell/evolution.html"]http://www.willamette.edu/~fruehr/haskell/evolution.html[/URL][/SIZE]