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Question Number 60635 by naka3546 last updated on 23/May/19

Answered by MJS last updated on 23/May/19

$$\mathrm{we}\mathrm{have}{6}^6=46656\mathrm{possible}\mathrm{results}$$$$33=$$$$666663\Rightarrow 6\mathrm{permutations}$$$$666654\Rightarrow 30\mathrm{permutations}$$$$666555\Rightarrow 20\mathrm{permutations}$$$$\Rightarrow \mathrm{probability}\mathrm{of}\mathrm{sum}=33\mathrm{is}\frac{56}{46656}=\frac{7}{5832}$$

Answered by mr W last updated on 23/May/19

$${(x+x^2+x^3+...+x^6)}^6=\frac{x^6{(1-x^6)}^6}{{(1-x)}^6}$$$$=x^6\left(\underset{k=0}{\sum ^6}{(-1)}^k{C}_k^6{x}^{6k}\right)\left(\underset{h=0}{\sum ^{\mathrm{\infty }}}{C}_5^{h+5}{x}^h\right)$$$$coef.of{x}^{33}=C_0^6{C}_5^{32}-C_1^6{C}_5^{26}+C_2^6{C}_5^{20}-C_3^6{C}_5^{14}+C_4^6{C}_5^8=56$$$$\Rightarrow thereare56possibilitiesthatthe$$$$sumis33.$$$$totalpossibilities=6^6$$$$\Rightarrow p=\frac{56}{6^6}\approx 0.12\mathrm{\%}$$

Commented by mr W last updated on 24/May/19

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