Case 1 -In how many ways the sum of upper faces of four distinct dices can be 6 ?

Answer

To find the number of ways the sum of the upper faces of four distinct dice can be 6, we can follow these steps:

### Step 1: Define the Variables
Let \( x_1, x_2, x_3, x_4 \) represent the numbers on the upper faces of the four distinct dice. We need to find the number of solutions to the equation:
\[ x_1 + x_2 + x_3 + x_4 = 6 \]
where \( 1 \leq x_i \leq 6 \) for \( i = 1, 2, 3, 4 \).

### Step 2: Change of Variables
To simplify the problem, we can perform a change of variables. Let:
\[ y_i = x_i - 1 \quad \text{for } i = 1, 2, 3, 4 \]
This transformation shifts the range of \( x_i \) from \( [1, 6] \) to \( [0, 5] \). Now, the equation becomes:
\[ (y_1 + 1) + (y_2 + 1) + (y_3 + 1) + (y_4 + 1) = 6 \]
which simplifies to:
\[ y_1 + y_2 + y_3 + y_4 = 2 \]
where \( 0 \leq y_i \leq 5 \).

### Step 3: Apply Stars and Bars Theorem
We can use the stars and bars theorem to find the number of non-negative integer solutions to the equation \( y_1 + y_2 + y_3 + y_4 = 2 \). The number of solutions is given by:
\[ \binom{n + k - 1}{k - 1} \]
where \( n \) is the total number of items to distribute (2 stars) and \( k \) is the number of variables (4 dice).

Here, \( n = 2 \) and \( k = 4 \), so we calculate:
\[ \binom{2 + 4 - 1}{4 - 1} = \binom{5}{3} \]

### Step 4: Calculate the Binomial Coefficient
Now we compute \( \binom{5}{3} \):
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Conclusion
Thus, the number of ways the sum of the upper faces of four distinct dice can be 6 is:
\[ \boxed{10} \]

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