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if 10! = 2^p3^q5^r7^s then...

if `10! = 2^p3^q5^r7^s` then

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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

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

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

Statement-1: The number of divisors of 10! Is 280. Statement-2: 10!= 2^(p)*3^(q)*5^(r)*7^(s) , where p,q,r,s in N.

Statement-1: The number of divisors of 10! Is 280. Statement-2: 10!= 2^(p)*3^(q)*5^(r)*7^(s) , where p,q,r,s in N.

Statement-1: The number of divisors of 10! Is 280. Statement-2: 10!= 2^(p)*3^(q)*5^(r)*7^(s) , where p,q,r,s in N.

Statement-1: The number of divisors of 10! Is 280. Statement-2: 10!= 2^(p)*3^(q)*5^(r)*7^(s) , where p,q,r,s in N.

Statement-1: The number of divisors of 10! Is 280. Statement-2: 10!= 2^(p)*3^(q)*5^(r)*7^(s) , where p,q,r,s in N.

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