JEE Advanced 2016 (10)

Probability Level 3

Compute the number of ordered quadruples of positive integers ( a , b , c , d ) (a,b,c,d) such that a ! ⋅ b ! ⋅ c ! ⋅ d ! = 24 ! . a! \cdot b! \cdot c!\cdot d! = 24!.

Source : OMO Spring 2016.
Try my set: JEE Advanced 2016
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Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Aryan Goyat
Apr 7, 2016

24! =23!(24) since 23 is prime it impossible to write it as product of factorials other than 1,0. so there are only few cases 24!.1!.1!.1!(4 cases) 23!.4!.1!.1!(12 cases) 23!.3!.2!.2!(12 cases)

6 Helpful 0 Interesting 0 Brilliant 0 Confused

So in general, what is the number of solutions of a! b! c! ... =k!?

Jun Arro Estrella - 5 years, 2 months ago

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