Question

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

Answer 1

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: $= 125$