The number of ways in which we can get a score of 11 by throwing three dice is...

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The number of ways in which we can get a score of 11 by throwing three dice is...

Mathematics
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HI!!
you familiar with dice?
yeah i am..

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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
none of the dice can show a one, but if one shows a 2, then it can be \[(2,5,6)\] or any combination of those there are 6 i believe \[(2,5,6),(2,6,5),(5,2,6),(5,6,2),(6,2,5),(6,5,2)\] or \(3!=6\) ways where one is a two
18,27,45,56 are the options
now lets move on to one showing a 3 \[(3,2,6)\] 6 of those \[(3,3,5)\] 3 of those \[(3,4,4)\] 3 of those guess that is it for one of them showing a 3
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 \[1,2,3,4,5,6,5,4,3,2,1\]
yeah count the number of ways for each for 3 dice the pattern is \[1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1\]
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
6 is not so big that we cannot list them \[(1,4,6),(1,6,4),(4,1,6),(4,6,1),(6,1,4),(6,4,1)\] six ways to do it
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
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and that is all if you go further, you will get repeats
add to get \[6+3+6+6+3+6=27\]
my approach here is to find the possibilities of getting 11 when the first die is a 1 then when a 2 etc eventually will add the possibilities
great @misty1212 thanks @triciaal
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I see 2 errors P(2) = 4 and P(6) = 8

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