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

- anonymous

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

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- misty1212

HI!!

- misty1212

you familiar with dice?

- anonymous

yeah i am..

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## More answers

- misty1212

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

- anonymous

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

- misty1212

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

- anonymous

18,27,45,56 are the options

- misty1212

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

- misty1212

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

- misty1212

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

- misty1212

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

- misty1212

keep counting in this way, you will get it

- misty1212

oh nvm we are done
just count them up

- anonymous

you made 12 cases

- misty1212

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

- misty1212

yeah count the number of ways for each
for 3 dice the pattern is
\[1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1\]

- misty1212

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

- misty1212

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

- anonymous

its confusing in counting cases

- misty1212

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

- misty1212

ok lets go slow

- misty1212

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

- misty1212

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

- misty1212

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

- anonymous

ok...

- misty1212

so lets do them in order

- misty1212

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

- misty1212

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

- misty1212

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

- triciaal

|dw:1433100469127:dw|

- misty1212

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

- misty1212

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

- triciaal

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

- anonymous

great @misty1212
thanks @triciaal

- misty1212

\[\color\magenta\heartsuit\]

- triciaal

|dw:1433101773240:dw|

- triciaal

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

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