A) 45
B) 48
C) 60
D) None of these
Correct Answer: C
Solution :
Let the boys gets a, b and c toys respectively. \[\therefore \] a + b + c = 14, a, b, c \[\ge \] 1 and a, b and c are distinct. Let a < b < c and \[{{x}_{1}}=a,{{x}_{2}}=b-a,{{x}_{3}}=a-b.\] So, \[3{{x}_{1}}+2{{x}_{2}}+{{x}_{3}}=14;{{x}_{1}},{{x}_{2}},{{x}_{3}}\ge 1\] \[\therefore \] The number of solutions = Coefficient of t14 in \[\{{{t}^{3}}+{{t}^{6}}+{{t}^{9}}+...)({{t}^{2}}+{{t}^{4}}+...)\]\[(t+{{t}^{2}}+...)\}\] = Coefficient of t8 in \[\{(1+{{t}^{3}}+{{t}^{6}}+....)(1+{{t}^{2}}+{{t}^{4}}+....)\] \[(1+t+{{t}^{2}}+....)\}\] = Coefficient of t8 in \[\{(1+{{t}^{2}}+{{t}^{3}}+{{t}^{4}}+{{t}^{5}}+2{{t}^{6}}+{{t}^{7}}+2{{t}^{8}})\] \[(1+t+{{t}^{2}}+....+{{t}^{8}})\}\] \[=1+1+1+1+1+2+1+2=10\] Since, three distinct numbers can be assigned to three boys in 3! ways. So, total number of ways = 10 x 3! = 60.
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