# Thread: N dice are thrown

1. ## 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?

2. ## Re: N dice are thrown

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?

3. ## Re: N dice are thrown

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?

4. ## Re: N dice are thrown

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).

5. ## Re: N dice are thrown

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

6. ## Re: N dice are thrown

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) $\binom{N}{3}\cdot 5^{N-3}$

3) $\binom{N}{2}\cdot(N-2)\cdot 4^{N-3}$

Now you tell us about the other two.

7. ## Re: N dice are thrown

Question 2

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

4.n!?

8. ## Re: N dice are thrown

Hello, ehpoc!

Do you know what binary coefficients are?

$n \ge 3$ 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: . $6^n$ possible outcomes.

(a) In how many ways can exactly 3 sixes turn up?
We want 3 Sixes and $n-3$ Others.

There are: . ${n\choose3}(1)^3(5)^{n-3}$ 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 $n$ Others: . ${n\choose0}(1^0)(5^n) \:=\:5^n$ ways.

1 six and $n-1$ Others: . ${n\choose1}(1^1)(5^{n-1}) \:=\:n\!\cdot\!5^{n-1}$ ways.

2 sixes and $n-2$ Others: . ${n\choose2}(1^2)(5^{n-2}) \:=\:\frac{n(n-1)}{2}\!\cdot\!5^{n-2}$ ways.

Less than 3 sixes: . $5^n + n\!\cdot\!5^{n-1} + \frac{n(n-1)}{2}\!\cdot\!5^{n-2} \:=\:\tfrac{1}{2}\,5^{n-2}(n^2 + 9n + 50)$ ways.

At least 3 sixes: . $6^n - \left[\tfrac{1}{2}\,5^{n-2}(n^2+9n + 50)\right]$ ways.

(c) In how many ways can exactly two 3's and one 4 turn up?

We want two 3's, one 4, and $n-3$ Others.

There are: . ${n\choose2,1,n\!-\!3}(1^2)(1^1)(5^{n-3})$ ways.

9. ## Re: N dice are thrown

Originally Posted by Soroban
[size=3]There are: . ${n\choose2,1,n\!-\!3}(1^2)(1^1)(5^{n-3})$ ways.
Here are a couple of comments (questions?)
Did you define ${n\choose2,1,n\!-\!3}~?$

Is it not ${n\choose2,1,n\!-\!3}(1^2)(1^1)({\color{blue}4^{n-3})}~?$

10. ## Re: N dice are thrown

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

11. ## Re: N dice are thrown

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: $\binom{N}{2}\binom{N-2}{1}(4^{N-3})$