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Suppose you are rolling two dice, one red and one white. Two rolls of the dice are considered equivalent if the dice sum to the same number. The dice are six-sided.

a) Give the partition induced by this equivalence relation.

b) How many equivalence relations exist on this set with the same size parts as the partition in part (a)?

Workings:

a. The partition would be numbers $1-6$ died on the red die and the numbers $1-6$ rolled on the white die.

b. There would be $36$ equivalence relations.

Roll a $1$ multiplied by the $6$ options on the other die to get $6$ multiply by $6$ again for the $6$ options on the original die.

I'm not sure if this is correct. Any help would be appreciated.

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Your answer to a) is wrong.

The question says "Two rolls of the dice are considered equivalent if the dice sum to the same number." The possible sums of the two dice range from $2$ through $12$. So one partition is all the possible rolls that sum to $2$, another with the sum $3$, and so on. This gives partitions of size $1,2,3,4,5,6,5,4,3,2,1$, as you note in your comment to this answer.

Now for part b). You need to find, out of a set of $36$, how many ways to get two sets of size $1$, two of size $2$, ..., two of size $5$, and one of size $6$. One way to do this is:

1) Find how many ways to choose $1$ from $36$, then $1$ from $35$. Multiply those two numbers, since they are independent, then divide by $2$ since you could have chosen either set first. (Note that the numbers here are independent, even though the choices themselves are not. You still multiple the numbers of choices.)

2) Find how many ways to choose $2$ from $34$, then $2$ from $32$. Multiply those two numbers, since they are independent, then divide by $2$ since you could have chosen either set first.

3-5) Do the same for choosing $3$, $4$, and $5$ from the numbers that are left.

6) You have only one choice to choose a set of $6$ from the remaining $6$ choices.

Then multiply those $6$ numbers together, since they are independent. There is no need to divide by anything, since sets of a given size are distinguishable from the other sets and cannot be interchanged.

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  • $\begingroup$ So there would be 11 partitions. b. The equivalence relations would be For sum of $2: 1+1 [1] $ For sum of $3: 1+2, 2+1 [2]$ " $4: 1+3, 2+2, 3+1 [3]$ " $5: 1+4, 2+3, 3+2, 4+1 [4]$ " $6: 1+5, 2+4, 3+3, 4+2, 5+1 [5]$ " $7: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1 [6]$ " $8: 2+6, 3+5, 4+4, 5+3, 6+2 [5]$ " $9: 3+6, 4+5, 5+4, 6+3 [4]$ " $10 4+6, 5+5, 6+4 [3]$ " $11: 5+6, 6+5 [2]$ " $12: 6+6 [1]$ $\endgroup$ – hockeynl Feb 2 '15 at 21:35
  • $\begingroup$ @hockeynl: That seems to be correct (except for the typo "104"). See my edited answer for help in part b). $\endgroup$ – Rory Daulton Feb 2 '15 at 22:05

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