Thread: N dice are thrown

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?

Originally Posted by ehpoc

I am asked N>=3 sets of dice are thrown. How many way can exactly 3 sixes turn up.

This is not a very clear description.
If N=5, are there 10 dice rolled (5 pair) or are there 5 single dice rolled?

LOL I even did a more clear description than the assignment did. Ignorant professor.

the EXACT question asked

N ≥ 3 dice are thrown. Express your answers to (a) to (d) in terms of binomial coefficients.

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.

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?

If it says "N ≥ 3 dice are thrown. Express your answers to (a) to (d) in terms of binomial coefficients" it means exactly that.
N dice are thrown. If N=5, then toss 5 dice.
Now list questions a) - d).

1. In how many ways can exactly 3 sixes turn up?
2. In how many ways can at least 3 sixes turn up?
3. In how many ways can exactly two 3's and one 4 turn up?
4. 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

Originally Posted by ehpoc

1. In how many ways can exactly 3 sixes turn up?
2. In how many ways can at least 3 sixes turn up?
3. In how many ways can exactly two 3's and one 4 turn up?
4. How many combinations of the dice are possible, in total?

This is a mixed set of problems.
1)

3)

Now you tell us about the other two.

Question 2

(N choose 3)*6^n-3?

4.n!?

Hello, ehpoc!

Do you know what binary coefficients are?

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?

There are: . possible outcomes.

(a) In how many ways can exactly 3 sixes turn up?

We want 3 Sixes and Others.

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.

Originally Posted by Soroban

[size=3]There are: . ways.

Here are a couple of comments (questions?)
Did you define

Is it not

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

Originally Posted by ehpoc

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

No indeed.
Choose two places for the two 3's.
Form the N-2 places left, choose one for the 4.
In the other N-3 places put any of the four, {1,2,5,6}, values.

ANSWER: