Rolling dice

Since the value of each dice is at least 1, the largest value on a dice is $10 - 5 = 5$ .

As such, the restriction that $x1 \leq 6$ is redundant, as it is automatically satisfied. We can look to the problem as the number of positive integer solutions of $x1+x2+x3+x4+x5+x6=10$ which is ${9 \choose 5} =126$ .

Since Maryann is rolling a 6-sided dice. The minimum number of the face of the dice is 1. So, we have to find the number of partitions of the number 10 into sum of 6 positive integers from the set {1, 2, 3, 4, 5, 6}.

The possible partitions of 10 into 6 positive integers from 1 to 6 are:

1+1+1+1+1+5

1+1+1+1+2+4

1+1+1+1+3+3

1+1+1+2+2+3

1+1+2+2+2+2

Thus, from each cases, we have a total of

$6!/5!$ + $6!/4!$ + $6!/(4!*2!$ )+ $6!/(3!*2!$ )+ $6!/(4!*2!$ )=6+30+15+60+15=126

Therefore, the answer is 126.

5,1,1,1,1,1=6 Ways

4,2,1,1,1,1=30 Ways

3,3,1,1,1,1=15 Ways

3,2,2,1,1,1=60 Ways

2,2,2,2,1,1=15 Ways

Total= 126 Ways

first we systematically write down all case where sum is 10 ant then we calculate their permutations (here the order of dice throws matters) :

_Case 1: $1+1+1+1+1+5 \Rightarrow \frac {6!}{5!} =6$ _

Case 2: $1+1+1+1+2+4 \Rightarrow \frac {6!}{4!} = 30$

Case 3: $1+1+1+1+3+3 \Rightarrow\frac {6!}{4! \cdot 2!} = 15$

Case 4: $1+1+1+2+2+3 \Rightarrow\frac {6!}{3! \cdot 2!} = 60$

Case 5: $1+1+2+2+2+2 \Rightarrow\frac {6!}{2! \cdot 4!} = 15$

(The formula used for permutation calculation is $\frac {n!}{a! \cdot b!}$ where $n=6$ and $a$ and $b$ are number of same thing)

Then the total is calculated which is $210$

If Maryann rolls a six-sided dice, there are six possible values for each dice roll. But the sum of the values has to be equal to ten. With this information, we can construct an equation relating the dice rolls:

$a + b + c + d + e + f = 10$

Where $a, b, c, d, e$ and $f$ represent, respectively, the value of the first dice roll, the value of the second dice roll... and the value of the sixth dice roll.

Since Maryann rolls a six-sided dice, the maximum value of each unknown is $6$ and the minimum value is $1$ . We want to find the positive integers that satisfy these conditions. Note that the equation above satisfies all the conditions of the problem, since the maximum value of each unknown, satisfying the "sum condition" (sum of $6$ positive integers equal to $10$ ) is $5$ , and the maximum number Maryann can get in a dice roll is $6$ .

Then, the unknown $a$ has to be a possitive integer and we can rewrite it as the sum $1 + u$ , where u is a non-negative integer. We do the same with $b, c, d, e$ and $f$ :

$a = 1 + u$

$b = 1 + v$

$c = 1 + w$

$d = 1 + x$

$e = 1 + y$

$f = 1 + z$

Therefore, the new equation will be:

$1 + u + 1 + v + 1 + w + 1 + x + 1 + y + 1 + z = 10$ $\Rightarrow$ $6 + u + v + w + x + y + z = 10$ $\Rightarrow$ $u + v + w + x + y + z = 4$

Now, if we find the possibile values for $u, v, w, x, y$ and $z$ , we will have the possible values for $a, b, c, d, e$ and $f$ . Using the counting technique "stars and bars", we can find the number of possible values. We know that we can associate each the solutions of this equation with a unique sequence of $0's$ and $1's$ . In this case, the sequences related would have $4$ $1's$ and $5$ $0's$ . The $0's$ can be combined in the $9$ "places" of the sequence, and there is only one way to combine the $1's$ , for each combination of $0's$ .

The number of combinations (and also the number of solutions to the equation) will be ${9 \choose 5} = 126$

Hence, there are $126$ ways Maryann can roll the dice satisfying the conditions of the problem.

So, the "stars and bars" technique can be reduced to a more simple formula. For example, the number of sotuins to the equation $a_{1} + a_{2} + a_{3} + ... + a{p} = n$ , where n is a positive integer and $a{1}, a{2}, a{3},... a{p}$ are non-negative integers, the number of solutions is given by ${(n + p - 1) \choose p-1}$ or ${(n + p - 1) \choose n}$

This is equivalent to the number of ways that $a + b + c + d + e + f = 10$ , given that each variable is an integer from 1 to 6. Notice that since each variable is at least 1, hence the sum of the other 5 variables is at least 5, so each variable is already at most 6.

Hence, the number of solutions is equal to the number of solutions such that $a + b + c + d + e + f = 10$ , given that each variable is an integer greater than or equal to 1.

There are ${ 10 - 1 \choose 6-1 } = 126$ ways to do so.

We have to divide 10 in 6 parts such that the least element is 1. By applying Stars and Bars technique, we see here, k=10 and n=6. Imagining k=10 stars placed on a line, there are k-1 = 9 gaps between them where bars can be placed as no bars can be placed at extreme left or extreme right of the line. Evidently, to divide these 10 stars into 6 positive groups, we just have to place only n-1 = 5 bars among the available gaps. Thus the result is 9C5 = 126

For there to be a sum of 10, exactly 10 dots from the die must appear. She throws it six times so there must be 6 groups of dots, such that the total number of dots is 10.

Therefore we have 5 intervals between the 6 groups and 9 spaces between the 10 dots, and we want to know the total number of ways the intervals can be arranged in the groups, now we can use 9 choose 5:

${9 \choose 5}=\frac{9!}{(9-5)!5!}=126$

Let $x_i + 1$ denote the face that shows up on the $i$ th roll. Clearly $x_i$ can take values between 0 and 5.

To solve the problem, we would like to know the total number of solutions to

$x_1 + 1 + x_2 + 1 + x_3 + 1 + x_4 + 1 + x_5 + 1 + x_6 + 1 = 10$

which simplifies to

$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 4$

By stars and bars , the total number of solutions is $\binom{6+4-1}{4} = 126$ .

Combination with Repetition Formula

C(n+r-1,r) = C(n+r-1,n-1)

where n equals to number of rolls and r equals to numbers of the sum of the dice/point

Assuming that each roll, we got at least 1 point.

[1][1][1][1][1][1]

Therefore, we only have 4 points left to be distributed to the 6 rolls.

By implementing the formula above, we can simply achieve the solution. C(6+4-1, 6-1) = C(9,5) = 126 ways