10 balls are distributed among 4 boxes. The probability for the first box always to contain 4 balls is

10 balls are distributed among 4 boxes

$n\left(S\right)=4^{10}$

E= the first box always to contain 4 balls 

4 balls can be selected  from 10 balls in ${ }^{10}{C}_4$ ways and can be kept in first box.

The remaining 6 balls to be distributed to other 3 boxes in 3⁶ ways

$n\left(E\right)={ }^{10}{C}_4{\left(3\right)}^6$

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{{ }^{10}{C}_4\times 3^6}{4^{10}}$