The number of ways of getting a sum of 16 whe | vedclass

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Question

(D) The number of ways to get a sum of $16$ with $4$ dice is the coefficient of $x^{16}$ in the expansion of $(x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^4$.T

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(D) The number of ways to get a sum of $16$ with $4$ dice is the coefficient of $x^{16}$ in the expansion of $(x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^4$.This is equal to the coefficient of $x^{16}$ in $[x(1-x^6)(1-x)^{-1}]^4 = x^4(1-x^6)^4(1-x)^{-4}$.We need the coefficient of $x^{12}$ in $(1-x^6)^4(1-x)^{-4}$.$(1-x^6)^4 = 1 - 4x^6 + 6x^{12} - \dots$$(1-x)^{-4} = 1 + \binom{4}{1}x + \binom{5}{2}x^2 + \dots + \binom{n+3}{3}x^n + \dots$The coefficient of $x^{12}$ is given by:$1 \cdot \binom{12+3}{3} - 4 \cdot \binom{6+3}{3} + 6 \cdot \binom{0+3}{3}$$= \binom{15}{3} - 4 \cdot \binom{9}{3} + 6 \cdot \binom{3}{3}$$= \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} - 4 	imes \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} + 6 	imes 1$$= 455 - 4 	imes 84 + 6$$= 455 - 336 + 6 = 125$.

The number of ways of getting a sum of $16$ when throwing a die four times is:

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