Question

The number of ways of getting a sum 16 on throwing a dice four times is __________.

Answer 1

Correct Answer: 125

Given expression:

\[ (x^1 + x^2 + \dots + x^6)^4 \]

Rewriting using binomial expansion:

\[ x^4 \left( \frac{1 - x^6}{1 - x} \right)^4 \]

Expanding:

\[ x^4 (1 - x^6)^4 (1 - x)^{-4} \]

Further expanding using binomial theorem:

\[ x^4 [1 - 4x^6 + 6x^{12} \dots ] \cdot (1 - x)^{-4} \]

Applying binomial expansion to each term:

\[ (x^4 - 4x^{10} + 6x^{16} \dots ) \cdot (1 + \binom{15}{12}x^{12} + \binom{9}{6}x^6 \dots) \]

Simplifying:

\[ (x^4 - 4x^{10} + 6x^{16}) \cdot \left(1 + \binom{15}{12}x^{12} + \binom{9}{6}x^6 \dots \right) \]

Computing coefficients:

\[ \binom{15}{3} - 4 \cdot \binom{9}{6} + 6 \]

Calculating values:

\[ = 35 \times 13 - 6 \times 8 \times 7 + 6 \]

Simplifying further:

\[ = 455 - 336 + 6 \]

Final result: \[ = 125 \]

Answer 2

Step 1: Define the situation
We are throwing a standard six-faced die four times. Each throw can result in an integer from 1 to 6. We are asked to find the total number of possible outcomes such that the sum of the four results equals 16.

Step 2: Formulate the equation
Let the four outcomes be \(x_1, x_2, x_3, x_4\), where \(1 \le x_i \le 6\).
We must find the number of integer solutions of \[ x_1 + x_2 + x_3 + x_4 = 16. \]

Step 3: Shift to nonnegative variables
Let \(y_i = x_i - 1\), then \(y_i \ge 0\) and \(y_i \le 5\). The equation becomes: \[ y_1 + y_2 + y_3 + y_4 = 16 - 4 = 12. \] We now count the number of nonnegative integer solutions of this equation with \(y_i \le 5.\)

Step 4: Ignore the upper bound first
Without the upper bound \(y_i \le 5\), the number of nonnegative integer solutions is \[ \binom{12 + 4 - 1}{4 - 1} = \binom{15}{3} = 455. \]

Step 5: Apply inclusion–exclusion for the upper bound
Let \(A_i\) be the condition \(y_i \ge 6\). Then substitute \(y_i' = y_i - 6\). The modified equation becomes: \[ y_1' + y_2 + y_3 + y_4 = 6. \] The number of solutions for one variable ≥ 6 is: \[ \binom{6 + 4 - 1}{4 - 1} = \binom{9}{3} = 84. \] Since there are 4 variables, total subtracted = \(4 \times 84 = 336.\)

For intersections where two variables ≥ 6, we subtract 12 and solve: \[ y_1' + y_2' + y_3 + y_4 = 0, \] which has exactly 1 solution for each such pair. The number of such pairs is \(\binom{4}{2} = 6\). Add back \(6 \times 1 = 6.\)

Step 6: Combine all using inclusion–exclusion
\[ \text{Valid solutions} = 455 - 336 + 6 = 125. \]

Step 7: Conclusion
Therefore, there are 125 ways to obtain a sum of 16 when a die is thrown four times.

Final answer

125