The number of ways of arranging 11 objects $A,B,C,D,E,F,\alpha ,\alpha ,\alpha ,\beta ,\beta$  so that every $\beta$  lie between two $\alpha$ (not necessarily adjacent) is $K\times 6!{\times }^{\text{11}}{\text{C}}_{\text{5}}$ , then K is ____.

There are three major ways  $\alpha \alpha \beta \beta \alpha ,\alpha \beta \beta \alpha \alpha ,\text{ and }\alpha \beta \alpha \beta \alpha$
Each major way has six empty spaces. The number of ways of putting letters at these empty spaces must be non-negative integer function of  $x_1+x_2+\mathrm{.....}+x_6=6$
 ${=}^{6+6-1}{C}_{6-1}{=}^{11}{C}_5$
No.of arrangements is  $3\left({ }^{11}{C}_5\right)6!$

or Number of ways to select 5 positions  among 11 in ${ }^{11}{C}_5$ ways in the selected places $\alpha \alpha \beta \beta \alpha ,\alpha \beta \beta \alpha \alpha ,\text{ and }\alpha \beta \alpha \beta \alpha$

can be arranged in 3  ways. in the remaining 6 places remaining 6 letters can be arranged in 6! ways.

No.of arrangements is  $3\left({ }^{11}{C}_5\right)6!$