The number of positive integral solutions is to be found

[ack]

I am preparing for an entrance and I came across this sum.The equation given is xyz=3000. we need to find how many positive integral solutions are there for x,y and z.
Please help.

 

[euroazn]

Let's factor 3000 into primes.
We are left with 2³*3*5³.

Now we are left with a very simple combinatorics question. Can you figure it out? Think about what items you are selecting and what you are placing them into.

 

[ack]

I already tried that.x can be 2^0,2^1,2^2or2^3 that is 4 ways ,2ways for 3and 4ways for 5.But what about y?
I thought of another method ,
using AM>GM,

x+y+z>=43 and max can be 3002 (when one of them is 3000 nd the other two 1 each.)
Then use multinomial theorem.Can this be done?

 

[euroazn]

I already tried that.x can be 2^0,2^1,2^2or2^3 that is 4 ways ,2ways for 3and 4ways for 5.But what about y?
I thought of another method ,
using AM>GM,

x+y+z>=43 and max can be 3002 (when one of them is 3000 nd the other two 1 each.)
Then use multinomial theorem.Can this be done?

You are vastly overthinking this.
Consider the following problem: we have 3 red balls, 3 blue balls, and a green ball. How many different ways can we allocate them between 3 buckets?

 

[ack]

Oh!...then the answer would be
9P7/(3!)^2.
thanks a lot!

 

[ack]

Oh!...then the answer would be
9P7/(3!)^2.
thanks a lot!

 

[ack]

Supposing we were to find all integral solutions?