If $10!=2^p{.3}^q{.5}^r{.7}^s$, then $\left(A\right)$ $2q=p$ $\left(B\right)$ $pqrs=64$ $\left(C\right)$ number of divisors of $10!$ is $280$ $\left(D\right)$ number of ways of putting $10!$ as a product of two natural numbers is $135$

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