Question

Four fair dice are rolled. Then the number of ways in which the sum of upper faces of four dices can be six, is

• A. \( 4 \)
• B. \( 10 \)
• C. \( 15 \)
• D. \( 24 \)
• E. \( 36 \)

Hint:

In dice-sum questions with small totals, convert the problem into an equation in positive integers. Then use stars and bars or pattern counting to get the answer quickly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem is equivalent to finding the number of positive integer solutions to an equation. Let the outcomes of the four dice be \(x_1, x_2, x_3, x_4\). We are looking for the number of solutions to the equation \(x_1 + x_2 + x_3 + x_4 = 6\), with the constraint that each \(x_i\) is an integer and \(1 \le x_i \le 6\).
Step 2: Key Formula or Approach:
Since the sum required (6) is small, the upper bound constraint \(x_i \le 6\) is automatically satisfied if \(x_i \ge 1\). So we only need to find the number of positive integer solutions.
We can transform this into a non-negative integer solution problem using a substitution. Let \(y_i = x_i - 1\), which implies \(y_i \ge 0\).
Substituting \(x_i = y_i + 1\) into the equation gives:
\((y_1+1) + (y_2+1) + (y_3+1) + (y_4+1) = 6\)
\(y_1 + y_2 + y_3 + y_4 = 2\)
This is a classic "stars and bars" problem. The number of non-negative integer solutions to \(y_1 + ... + y_k = n\) is given by \(C(n+k-1, k-1)\).
Step 3: Detailed Explanation:
We need to find the number of non-negative integer solutions for \(y_1 + y_2 + y_3 + y_4 = 2\).
Here, \(n=2\) (the sum, or "stars") and \(k=4\) (the number of variables, or "bins").
Using the stars and bars formula:
\[ \text{Number of ways} = C(n+k-1, k-1) = C(2+4-1, 4-1) = C(5, 3) \] Now, we calculate the combination:
\[ C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 \] Alternative Method (Listing Partitions):
We can also list the possible combinations of numbers that sum to 6:
1. 3, 1, 1, 1: The number 3 can be on any of the 4 dice. This gives \(\frac{4!}{3!1!} = 4\) ways.
2. 2, 2, 1, 1: We need to arrange two 2s and two 1s. This gives \(\frac{4!}{2!2!} = \frac{24}{4} = 6\) ways.
Total number of ways = 4 + 6 = 10.
Step 4: Final Answer:
The number of ways in which the sum of the upper faces can be six is 10. Therefore, option (B) is correct.