A) \[^{8}{{C}_{3}}\]
B) 21
C) \[{{3}^{8}}\]
D) 5
Correct Answer: B
Solution :
[b] Required number of ways = coefficient of \[{{x}^{8}}\] in \[{{(x+{{x}^{2}}+..{{x}^{6}})}^{3}}\] [\[\because \] Each box can receive minimum 1 and maximum 6 balls] = coeff of \[{{x}^{8}}\] in \[{{x}^{2}}{{(1+x+{{x}^{2}}+...+{{x}^{5}})}^{3}}\] = coeff of \[{{x}^{5}}\] in \[{{\left( \frac{1-{{x}^{6}}}{1-x} \right)}^{3}}\] = coeff of \[{{x}^{5}}\] in \[{{(1-x)}^{-3}}\] = coeff of \[{{x}^{5}}\] in \[(1{{+}^{3}}{{C}_{1}}x{{+}^{4}}{{C}_{2}}{{x}^{2}}+...)\] \[{{=}^{7}}{{C}_{5}}=21\]
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