A) 70
B) 35
C) 64
D) 192
Correct Answer: B
Solution :
Required number of ways = Coefficient of \[{{x}^{16}}\] in \[{{({{x}^{3}}+{{x}^{4}}+{{x}^{5}}+....+{{x}^{7}})}^{4}}\] = Coefficient of \[{{x}^{16}}\] in \[{{x}^{12}}\]\[{{(1+x+{{x}^{2}}+....{{x}^{4}})}^{4}}\] = Coefficient of \[{{x}^{16}}\] in \[{{x}^{12}}{{(1-{{x}^{5}})}^{4}}{{(1-x)}^{-4}}\] = Coefficient of \[{{x}^{4}}\] in \[{{(1-{{x}^{5}})}^{4}}{{(1-x)}^{-4}}\] = Coefficient of \[{{x}^{4}}\] in \[(1-4{{x}^{5}}+...)\]\[\left[ 1+4x+....+\frac{(r+1)\,(r+2)\,(r+3)}{3\,!}\,{{x}^{r}} \right]\] = \[\frac{(4+1)(4+2)(4+3)}{3\,!}=35\]. Aliter: Remaining 4 rupees can be distributed in \[^{4+4-1}{{C}_{4-1}}\] i.e., 35 ways.
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