permutations

[mfgoode]

Originally Posted by davieddy

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

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

Mally

[davieddy]

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

[mfgoode]

Originally Posted by davieddy

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

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

[Wacky]

Originally Posted by mfgoode

:
I understand that both (60!/4!)^15
Where is the mistake Sir ?

Originally Posted by davieddy

we get 60!/4!^15.

Mally,

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

[davieddy]

Originally Posted by mfgoode

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

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

[mfgoode]

Originally Posted by Wacky

Mally,

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

: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

[davieddy]

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)

[Richard Cameron]

Originally Posted by mfgoode

: Just out of curiousity I computed the number. It has 11 trailing zeroes!

Mally

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

[davieddy]

Originally Posted by mfgoode

: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

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

David

[mfgoode]

Originally Posted by davieddy

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)

Thank you Davie, thats a new one for me.

Easy to remember too -- Bed Mistress as Sweetheart!

Mally

[mfgoode]

Originally Posted by Richard Cameron

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

:

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