Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

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

OpenStudy (misty1212):

HI!!

OpenStudy (misty1212):

you familiar with dice?

OpenStudy (anonymous):

yeah i am..

OpenStudy (misty1212):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

18,27,45,56 are the options

OpenStudy (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

OpenStudy (misty1212):

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

OpenStudy (misty1212):

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

OpenStudy (misty1212):

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

OpenStudy (misty1212):

keep counting in this way, you will get it

OpenStudy (misty1212):

oh nvm we are done just count them up

OpenStudy (anonymous):

you made 12 cases

OpenStudy (misty1212):

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

OpenStudy (misty1212):

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

OpenStudy (misty1212):

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

OpenStudy (misty1212):

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

OpenStudy (anonymous):

its confusing in counting cases

OpenStudy (misty1212):

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

OpenStudy (misty1212):

ok lets go slow

OpenStudy (misty1212):

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

OpenStudy (misty1212):

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

OpenStudy (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

OpenStudy (anonymous):

ok...

OpenStudy (misty1212):

so lets do them in order

OpenStudy (misty1212):

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

OpenStudy (misty1212):

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

OpenStudy (misty1212):

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

OpenStudy (triciaal):

|dw:1433100469127:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!