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: Translate the dice problem into an equation.
Let the numbers appearing on the four dice be \[ x_1,x_2,x_3,x_4 \] Since each die shows at least \( 1 \), we have \[ x_1+x_2+x_3+x_4=6 \] where \[ x_i \geq 1 \] for each \( i=1,2,3,4 \).

Step 2: Convert to a non-negative integer equation.
Let \[ y_1=x_1-1,\quad y_2=x_2-1,\quad y_3=x_3-1,\quad y_4=x_4-1 \] Then each \( y_i \geq 0 \), and the equation becomes \[ (y_1+1)+(y_2+1)+(y_3+1)+(y_4+1)=6 \] \[ y_1+y_2+y_3+y_4=2 \]

Step 3: Use the stars and bars formula.
Now we need the number of non-negative integer solutions of \[ y_1+y_2+y_3+y_4=2 \] The number of solutions is given by \[ {}^{n+r-1}C_{r} \] where \( n=4 \) variables and \( r=2 \) units.
So, the number of solutions is \[ {}^{4+2-1}C_2={}^{5}C_2 \]

Step 4: Evaluate the combination.
\[ {}^{5}C_2=\frac{5 \cdot 4}{2 \cdot 1}=10 \] Hence, there are \( 10 \) ordered outcomes of four dice whose sum is \( 6 \).

Step 5: Verify by direct pattern method.
Let us check possible distributions of sum \( 6 \) among \( 4 \) dice, each at least \( 1 \):
The only possible patterns are: \[ 3,1,1,1 \] and \[ 2,2,1,1 \] There are no other possibilities because the minimum sum of four dice is already \( 4 \).

Step 6: Count arrangements of these patterns.
For \[ (3,1,1,1) \] the number of arrangements is \[ \frac{4!}{3!}=4 \] For \[ (2,2,1,1) \] the number of arrangements is \[ \frac{4!}{2!2!}=6 \] So total number of ways is \[ 4+6=10 \] This confirms our earlier result.

Step 7: Final conclusion.
Therefore, the number of ways in which the sum of upper faces of four dice is \( 6 \) equals \[ \boxed{10} \] Hence, the correct option is \[ \boxed{(2)\ 10} \]