A) \[\frac{1}{2}\]
B) \[\frac{7}{15}\]
C) \[\frac{2}{15}\]
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
[b] Total number of ways of arranging the balls \[=\frac{10!}{3!7!}=120\] Favourable cases, \[x{{B}_{1}}y{{B}_{2}}z{{B}_{3}}t\] If \[x,y,z\] and t be the number of white balls to be Place as shown above then \[0\le x,t\le 5\] \[1\le y,z\le 6\] \[\therefore \] Number of favourable cases = coefficient of \[{{x}^{7}}\] in \[{{(1+x+{{x}^{2}}+...+{{x}^{5}})}^{2}}{{(x+{{x}^{2}}+...+{{x}^{6}})}^{2}}\] = coeff. of \[{{x}^{7}}\] in \[{{x}^{2}}{{(1+x+{{x}^{2}}+...+{{x}^{5}})}^{4}}\] = coeff. of \[{{x}^{5}}\] in \[{{\left( \frac{1-{{x}^{6}}}{1-x} \right)}^{4}}=coeff.\,\,of\,\,{{x}^{5}}\,\,in\] \[{{(1-x)}^{-4}}\] \[{{=}^{8}}{{C}_{5}}=56\] \[\therefore \] Desired probability \[=\frac{56}{120}=\frac{7}{15}\]
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