Number of possibilities for throwing 2 identical dice?

11 12 13 14 15 16 22 23 24 25 26 33 34 35 36 44 45 46 55 56 66 tital 21

[math]\frac{36-6}{2}+6[/math]

Can someone translate this into a stars and bars problem?

File: 1487451162908.gif (537KB, 480x270px) Image search: [Google]

537KB, 480x270px

>>8691377
Post the actual problem.

>>8691377
6+5+4+3+2+1

File: dice.png (1KB, 295x119px) Image search: [Google]

1KB, 295x119px

>>8691377
here's a visualization

>>8691377
(6 choices for die 1)*(5 choices for die 2)/(2 ways of arranging 2 distinct dice)
+
(6 choices for die 1)*(1 choice for die 2)/(1 way of arranging 2 identical dice)

You have 6*6 possibilities. 36.

But if you get 1;1 or 2;2 or 3;3 ... it's the same thing. So you subtract those 6 cases.

36 - 6 = 30

But you also have that 1;2 and 2;1 are the same. Just like 3;6 and 6;3 are the same.
So you must subtract (36/2)/2 = 9

30 - 9 = 21

>>8691377
Combos of two dice without repetition:
[math]
= \frac{(6+2-1)!}{2!5!}\\
= \frac{7!}{2!(6-1)!}\\
= \frac{6*7}{2*1}\\
= 21
[/math]

>>8691487
meant to edit that to illustrate the (6-1)! in the first part and then simplify. The point is the formula is
[math]
\frac{(n+r-1)!}{r!(n-1)!}
[/math]

where n is the set you choose from, r is the number of items you choose.

>>8691479
>So you must subtract (36/2)/2
Can you explain this further? I'm counting more than 9 of these cases:
1;2 1;3 1;4 1;5 1;6 These can all happen twice, so subtract out 5
2;3 2;4 2;5 2;6 These all happen twice, so subtract out 4
3;4 3;5 3;6 These all happen twice, so subtract out 3.
Already we are subtracting more than 9, and we haven't even gotten to 4;5 4;6 or 5;6 which can all happen twice.

>>8691492
I fucked up.