How many different 7 letter combinations can be drawn from a fresh bag?

A fresh bag has 98 tiles and 2 blanks.

How many can be drawn with only the 98 tiles and 0 blanks?

If all the tiles were different it would be an easier problem, but somehow the varying frequencies of the different tiles seem to complicate the issue.

Let Nk be the number of tiles for the kth letter, k = 1 to 26

Total Ways for 7 tiles to be drawn = Sum over J { C(N1,J1)*C(N2,J2)* . . . *C(N26,J26)}

where the summation over J is such that J1+J2 + . . . +J26 = 7 and N1+N2+ . . . +N26 = 98 and all J >=0.

Oh, don’t give me the N values and expect me to calculate this.

Also, I don’t guarantee that this is the best way and I know it’s not the only way.

Well a good starting place would be to post the numbers of each tile.

I counted 3,199,724 unique tile combinations for the English version.

First make a table of (T,L) where L letters have T or more tiles.


Now form all of the partitions of the number 7. That is, the number of ways to sum positive integers to form 7 without regard to order. There are 15 such partitions as is well known. List these using notation (T,L) where we take T tiles from each of L letters.


Now count the number of combinations in each of these partitions using the first table and sum:

4 +
7*26 +
7*21 +
7*C(26,2) +
11*11 +
11*21*25 +
11*C(26,3) +
C(12,2)*25 +
12*C(21,2) +
12*21*C(25,2) +
12*C(26,4) +
C(22,3)*24 +
C(22,2)*C(25,3) +
22*C(26,5) +
C(27,7)

= 3,199,724


The first 2 elements of the first table change to 26 and 21. Then the calculation becomes:

4 +
7*25 +
7*20 +
7*C(25,2) +
11*11 +
11*20*24 +
11*C(25,3) +
C(12,2)*24 +
12*C(20,2) +
12*20*C(24,2) +
12*C(25,4) +
C(21,3)*23 +
C(21,2)*C(24,3) +
21*C(25,5) +
C(26,7)

= 2,484,184

That's just C(98,7). We want unique combinations rather than total combinations. So if all 7 letters are letter number 1 so J1=7 for example, that only counts as 1, not C(N1,7). Note that in the C(n,k) terms in my formula, n is a number of distinct letters rather than the number of tiles for a given letter. We only care about the number of tiles for a given letter insofar as it is large enough for that letter to be included in particular partitions.

I thought there was probably an easier way to get an answer to a question that wasn't asked.

I went ahead and did the second part ignoring the 2 spaces and added it to the original post. I just had to decrease all the numbers in the 20s by 1. The answer isn't that far from the number of 5 card poker hands.

Notice that if there were 7 or more copies of every letter, the answer wouldn't depend on the distribution of letters at all. It would be the same as if there were infinitely many of each letter. We can use this as a sanity check on the answer to get an upper bound. In the case where we have 27 letters (including the space) the answer would simply be C(27+7-1,7). It's the same as the number of distinguishable ways to put 7 balls in 27 buckets numbered 1 to 27 where the balls are indistinguishable. This is the same as the number of ways to put 7 balls in 33 buckets if you can only put 1 ball per bucket. DUCY? This number is 4,272,048 which is indeed bigger than our answer of 3,199,724. For the case of 26 letters and no space, the upper bound is C(26+7-1,7) = 3,365,856 which again is bigger than our answer of 2,484,184. These bounds aren't real tight, but at least we can be sure that the answer can't be any bigger than that.

We can also use the partition method to arrive at the upper bound of 4,272,048 which verifies that basic method. We would start with the first table having all entries of L=27 down to T=7 since all letters would have at least 7 copies. Then the calculation would be:

27 +
27*26 +
27*26 +
27*C(26,2) +
27*26 +
27*26*25 +
27*C(26,3) +
C(27,2)*25 +
27*C(26,2) +
27*26*C(25,2) +
27*C(26,4) +
C(27,3)*24 +
C(27,2)*C(25,3) +
27*C(26,5) +
C(27,7)

= 4,272,048

The same as C(27+7-1,7).

Good stuff Bruce, thanks.