A batsman scores exactly a century by hitting fours and sixes in twenty consecutive balls. In how many different ways can he do it if some balls may not yield runs and the order of boundaries and Over boundaries are taken into account?

Let the batsman hit $x$ fours, $y$ sixes and let $z$ balls may not yield runs. Then we have,
$4x+6y+0z=\mathrm{100....}(1)$
Where
$x+y+z=\mathrm{20...}(2)$
From Eqs. (1) and (2),
$4(20-y-z)+6y=100$
$\Rightarrow 2y-4z=20$
$\Rightarrow y-2z=10$
$\Rightarrow y=10+2z$
Hence, $z$ can be $0,1,2,3$ as any of $x,y,z$ cannot exceed $20$. So, we have following type of distribution of runs:

Number of fours, $x$	Number of sixes, $y$	Number of zeros, $z$	Number of ways of arranging $x,y$ and $z$
$10$	$10$	$0$	$\frac{20!}{10!10!}$
$7$	$12$	$1$	$\frac{20!}{7!12!}$
$4$	$14$	$2$	$\frac{20!}{4!14!2!}$
$1$	$16$	$3$	$\frac{20!}{16!3!}$

Hence, the total number of ways is
$20!\left(\frac{1}{16!3!}+\frac{1}{14!4!2!}+\frac{1}{12!7!}+\frac{1}{10!10!}\right)$.