Symbiosis National Aptitude Test SNAP 2019 Paper

${\left(5!\right)}^{5!}+{\left(10!\right)}^{10!}$ + ${\left(50!\right)}^{50!}+{\left(100!\right)}^{100!}$
+ ${\left(49!\times 5\times 10\right)}^{50!}+{\left(99!\times 100\right)}^{100!}$
= ${\left(12\times 10\right)}^{5!}+{\left(9!\times 10\right)}^{10!}$ + ${\left(49!\times 5\times 10\right)}^{50!}+{\left(99!\times 100\right)}^{100!}$
Now, ${\left(10\right)}^{5!}$ = ${\left(10\right)}^{120}$ is a number that have 120 zeroes
${\left(10\right)}^{10!}$ is a number that have more than 120 zeroes at the end.
${\left(10\right)}^{50!}$ is a number that have more than 120 zeroes at the end.
and ${\left(100\right)}^{100!}$ is a number that have more than 120 zeroes at the end.
So, number of zeroes at the end of the sum = 120.