Question 875472: Hello. I have a question regarding the number of combinations in the game Qwirkle. In Qwirkle, each chip has a shape and a color. There are six shapes and six colors, resulting in a total of 36 types of chips. There are three replications of each type. Thus, there are 108 chips in total. Six chips are initially drawn at random. So I am interested in finding the unique combinations of chips that can initially be drawn. Knowing this number will enable me to compute the probability of various outcomes. Here is the approach I took:
1) assume chips of the same type are indistinguishable (e.g. a blue square and another blue square are the same)
2) enumerate integer partitions with the restriction that the integers must sum to six and a single integer cannot exceed three (because there are only three of each type).
3) count the number of integers in each partition, K, and compute 36 choose K
4) sum the combinations from step 3
Partition 1: 3,3
Partition 2: 3,1,1,1
Partition 3: 2,1,1,1,1
Partition 4: 1,1,1,1,1,1
Partition 5: 2,2,2
Partition 6: 2,2,1,1
Partition 7: 3,2,1
Answer: 36 choose 2 + 36 choose 4 + 36 choose 5 + 36 choose 6 + 36 choose 3 + 36 choose 4 + 36 choose 3
Is my reasoning correct?
Thank you!
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
I think it's this way for the 7 partitions:
------------------------------------------------
Partition 1: 3,3
Choose the types for the 2 triples in 36C2 ways.
That's 36C2 = 630 ways.
------------------------------------------------
Partition 2: 3,1,1,1
Choose the type for the triple in 36C1 ways
Choose the types for the 3 singles in 35C3 ways
That's 36C1×35C3 = 36×6545 = 235620 ways
[Notice here that if you choose the 3 singles first in 36C3 ways,
then the triple in 33C1 ways,
that would be 36C3×33C1 = 7140×33 = 235620 ways, which shows that
it doesn't matter.]
------------------------------------------------
Partition 3: 2,1,1,1,1
Choose the type for the double in 36C1 ways
Choose the types for the 4 singles in 35C4 ways
That's 36C1×35C4 = 36×52360 = 1884960 ways
[Notice here that if you choose the 4 singles first in 36C4 ways,
then choose the triple in 32C1 ways,
that would be 36C4×32C1 = 58905×32 = 1884960 ways, which shows that
it doesn't matter]
------------------------------------------------
Partition 4: 1,1,1,1,1,1
Choose the types for the 6 singles in 36C6 ways
That's 36C6 = 1947792 ways
------------------------------------------------
Partition 5: 2,2,2
Choose the types for the 3 doubles in 36C3 ways
That's 36C3 = 7140 ways.
------------------------------------------------
Partition 6: 2,2,1,1
Choose the types for the 2 doubles in 36C2 ways
Choose the types for the 2 singles in 34C2 ways
That's 36C2×34C2 = 630×561 = 353430 ways.
------------------------------------------------
Partition 7: 3,2,1
Choose the type for the triple in 36C1 ways
Choose the type for the double in 35C1 ways
Choose the type for the single in 34C1 ways
That's 36C1×35C1×34C1 = 36×35×33 = 42840 ways.
------------------------------------------------
630 + 235620 + 1884960 + 1947792 + 7140 + 353430 + 42840 = 4472412 total.
Edwin
|
|
|
