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If 10! =2^(p).3^(q).5^(r).7^(s), then -...

If `10! =2^(p).3^(q).5^(r).7^(s)`, then -

A

2q =p

B

pqrs =64

C

number of divisors of `10!` is 280

D

number of ways of putting 10 ! As a product of two natural numbers is 135

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Knowledge Check

  • (10!)=(2)^p.(3)^q.(5)^r.(7)^8 then

    A
    2q=p
    B
    pqrs=64
    C
    number of divisors of `10!` is 280
    D
    number of ways of putting `10!` as a product of two natural numbers is 135

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