I am asked N>=3 sets of dice are thrown. How many way can exactly 3 sixes turn up.
I did the following
for a single set of dice....
dice1+dice2=6 how many solutions?
(5 choose 1)/2ceiling function??? which is 3 ways
so for 3 dice to roll 6 there are...
3x3x3
then the other n-3 dice can roll anything but a six.
so there are 18 unique ways to roll a set of dice
18-3=15 -the total ways to roll a dice with no 6.
so I get...
(3)(3)(3)(n-3)(15) ways.
Now my big problem is how do I express this answer in terms of binomial co-efficient?
LOL I even did a more clear description than the assignment did. Ignorant professor.
the EXACT question asked
I inferred that, that meant n sets f dice 2n die in total. Since this is kind of how we have been doing this in class.N ≥ 3 dice are thrown. Express your answers to (a) to (d) in terms of binomial coefficients.
But now I am not sure at all. I am waiting for an email from the prof. Either way the most confusing part is how do I express my answer as a binomial coefficient?
- In how many ways can exactly 3 sixes turn up?
- In how many ways can at least 3 sixes turn up?
- In how many ways can exactly two 3's and one 4 turn up?
- How many combinations of the dice are possible, in total?
My biggest confusion is mainly in the part of expressing my answer as a niary coefficient
Hello, ehpoc!
Do you know what binary coefficients are?
There are: . possible outcomes.dice are thrown.
Express your answers for (a) to (d) in terms of binomial coefficients.
(d) How many combinations of dice dice are possible in total?
We want 3 Sixes and Others.(a) In how many ways can exactly 3 sixes turn up?
There are: . ways.
(b) In how many ways can at least 3 sixes turn up?
The opposite of "at at least 3 sixes" is: .0 sixes or 1 six or 2 sixes.
0 sixes and Others: . ways.
1 six and Others: . ways.
2 sixes and Others: . ways.
Less than 3 sixes: . ways.
At least 3 sixes: . ways.
(c) In how many ways can exactly two 3's and one 4 turn up?
We want two 3's, one 4, and Others.
There are: . ways.
Ummm I think that answer for B is a little crazy.
(Nc3) gets our three sixes
then the other dice can be anything
6^n-3
Does this not make sense?
Also.
Plato, for you answer for c.
you have N choose 2
Is it not N choose 3 still? we still need to choose 3 dice. Two 3's and a 4?
(nC3)(4)^n-2