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HI!!

you familiar with dice?

yeah i am..

ok i guess we got a bunch of counting to do then because you have 3 dice total, not the usual 2

yep two dices would be easy.. but there's 3

18,27,45,56 are the options

ready to move on to 4? we have to make sure we don't count anything twice

oh i made a mistake
one could show a 1 lets count those too

\[(1,5,5)\] 3 of those and
\[(1,4,6)\] 6 of those

keep counting in this way, you will get it

oh nvm we are done
just count them up

you made 12 cases

for two dice the pattern is
\[1,2,3,4,5,6,5,4,3,2,1\]

for rolling respectively
\[3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\]

notice that
\[(1,4,6)\] where all the numbers are different has \(3!=6\) possible ways to do it

its confusing in counting cases

whereas \((1,5,5)\) has only 3 possible ways
\[(1,5,5),(5,1,5),(5,5,1)\]

ok lets go slow

you got one possibility is \((1,4,6)\) right because they add up to 11

but you have \(3!=6\) permutations of those three numbers

ok...

so lets do them in order

\[(1,4,6)\] 6 choices
\[(1,5,5)\] 3 choices

\[(2,3,6)\] 6 choices
\[(2,4,5)\] 6 choices

\[(3,3,5)\]3 choices
\[(3,4,6)\] 6 choices

|dw:1433100469127:dw|

and that is all
if you go further, you will get repeats

add to get \[6+3+6+6+3+6=27\]

great @misty1212
thanks @triciaal

\[\color\magenta\heartsuit\]

|dw:1433101773240:dw|

I see 2 errors P(2) = 4 and P(6) = 8