Let N be the number of ways of distributing 52 identical balls into...

Let $i^{\mathrm{h}}$ box has $6 \lambda_{\mathrm{i}}+\mu_{\mathrm{i}}$ balls where $\lambda_i, \mu_i \in w$ and $\mu_i \leq 5$. Also, all $\mu_{\mathrm{i}}$ 's are distinct. $\because \quad 6\left(\lambda_1+\lambda_2+\lambda_3+\lambda_4\right)+\left(\mu_1+\mu_2+\mu_3+\mu_4\right)=52 \text { and } \mu_1+\mu_2+\mu_3+\mu_4 \in\{6,7,8,9,10,11,12,13,14\}$ Hence, only one possibility is there $$\begin{array}{ll} & \lambda_1+\lambda_2+\lambda_3+\lambda_4=7 ...(i)\\ \text { and } & \mu_1+\mu_2+\mu_3+\mu_4=10...(ii) \end{array}$$ Equation (ii) has only three solution sets, which are $(5,4,1,0),(5,3,2,0)$ and $(4,3,2,1)$. Equation (i) has total ${ }^{10} C_3$ solutions and ${ }^4 C_1 \cdot{ }^9 C_2$ solutions when any $\lambda_i$ is zero. So, $N=3$\left({ }^{10} C_3 \cdot(4!)\right)-2\left({ }^4 C_1 \cdot{ }^9 C_2(3!)\right)=6912 \therefore \quad a=69 \text { and } b=12 .