Counting primes in Scrabble racks

[retina]

Take a standard English set of Scrabble tiles:

 0 points: Blank ×2
 1 point:  E ×12, A ×9, I ×9, O ×8, N ×6, R ×6, T ×6, L ×4, S ×4, U ×4
 2 points: D ×4, G ×3
 3 points: B ×2, C ×2, M ×2, P ×2
 4 points: F ×2, H ×2, V ×2, W ×2, Y ×2
 5 points: K ×1
 8 points: J ×1, X ×1
10 points: Q ×1, Z ×1

Count the number of unique racks you can draw, from a full bag, for draws of 1 tile through 50 tiles. Factorise, and post the largest prime found among all draw sizes.

e.g.

 1 tile draw - 27 unique racks = (3*3*3)
 2 tile draw - ?? unique racks = (a*b*...*c)
...
50 tile draw - ?? unique racks = (x*y*...*z)

And post the largest number among a...z (there is only one unique answer).

BTW: The point value of each tile means nothing for this puzzle, you can safely ignore it.

[edit]
Just to make sure this is clear.

Each new draw is started with a full set of 100 tiles.
Once you have started drawing, the tiles are not replaced into the bag, they put on the rack.

So, for 1 tile draws:

Start with 100 tiles, draw an "A". That counts as 1 possible draw set.
Then start again with 100 tiles, draw a "B". Now we have 2 possible sets.
...
Until lastly we draw the blank, to make 27 unique sets of draws.

Permutations are not counted as unique, "ABC"="CBA"="BCA"="ACB"="BAC"="CAB" is only counted as one set.
[/edit]

[Brian-E]

My calculations for drawing 2 tiles and 3 tiles are below. The method I use is too laborious for much further progress to be feasible. Plus I'm not sure if I'm not making a mistake.

2 tiles:
22 ways if both tiles are the same (because 5 of the letters cannot be used)
27*26/2! if they are different
Total: 373 (prime)

3 tiles:
12 ways if all three tiles are the same
2*5*22 ways if two of the tiles are the same
27*26*25/3! if all three are different
Total: 3157 (=7*11*41)

[Brian-E]

My solution for the 3 tiles was wrong, correction:

3 tiles:
12 ways if all three tiles are the same
22*26 if two of them are the same (these two cannot be any of the 5 single letters)
27*26*25/3! if all three are different
Total: 3509 (=11^2*29)

Going further seems to require lots of effort in dealing with all the different possibilities. I would undoubtably make more mistakes if I tried. Does anyone have a better method?

[retina]

Originally Posted by Brian-E

My solution for the 3 tiles was wrong, correction:

3 tiles:
12 ways if all three tiles are the same
22*26 if two of them are the same (these two cannot be any of the 5 single letters)
27*26*25/3! if all three are different
Total: 3509 (=11^2*29)

Correct for 2 and 3 tiles

Originally Posted by Brian-E

Going further seems to require lots of effort in dealing with all the different possibilities. I would undoubtably make more mistakes if I tried. Does anyone have a better method?

The method for this and the recent darts puzzle are closely related. Anything more than 1ms CPU time to get all 50 values and you have some room for improvement.

[Brian-E]

Extended to 4 and 5 tiles but I'm not succeeding in generalising the method so that it can be programmed.

2 tiles: 373 (prime)
3 tiles: 3509 (11^2*29)
4 tiles: 18104 (2^3*31*73)
5 tiles: 148024 (2^3*18503)

[retina]

Originally Posted by Brian-E

4 tiles: 18104 (2^3*31*73)
5 tiles: 148024 (2^3*18503)

Doesn't look right to me.

[Brian-E]

Originally Posted by retina

Doesn't look right to me.

My way of tackling this problem is much too laborious and error-prone. I found an error in my working of both 4- and 5-tiles. Here are corrected versions, but the chance that they are now error-free is still not good.

4 tiles:
All 4 the same letter: 11 ways
3 the same, 1 of another: 12*26 ways
"2+2": 22*21/2! ways
"2+1+1": 22*26*25/2! ways (forgot this possibility before)
"1+1+1+1": 27*26*25*24/4! ways
Total: 25254 (2*3^2*23*61)

5 tiles:
"5 all the same": 7 ways
"4+1": 11*26 ways
"3+2": 12*21 ways (mistakenly divided this by 2 before, but the ordering is important here)
"3+1+1": 12*26*25/2! ways
"2+2+1": 22*21*25/2! ways
"2+1+1+1": 22*26*25*24/3! ways
"1+1+1+1+1": 27*26*25*24*23/5! ways
Total: 148150 (2*5^2*2963)

Generalising this method and programming it doesn't really appeal to me. I am wondering is there is a more elegant way of counting.

[retina]

Originally Posted by Brian-E

4 tiles:
Total: 25254 (2*3^2*23*61)

5 tiles:
Total: 148150 (2*5^2*2963)

This looks correct now.

Originally Posted by Brian-E

Generalising this method and programming it doesn't really appeal to me. I am wondering is there is a more elegant way of counting.

So far, you have done it the hard way, and I am impressed that you took it all the way to 5 tiles.

[ckdo]

I'll take a shot at answering the original question: 0xC75407108F5D?

[retina]

Originally Posted by ckdo

I'll take a shot at answering the original question: 0xC75407108F5D?

Correct.

[ckdo]

Originally Posted by retina

Correct.

Boy am I glad. Getting the 50 values certainly is a millisecond task, but factoring them takes me somewhat over a second. I'll try to work on this.