The no. of +ve integral solutions of x₁x₂x₃x₄=210 is=?

Hi Basit,

Fist factorise 210 = 2*3*5*7.

Now, this has no "1".

When you want to have one "1". Then from the above 4 factors, pick two and club it. Which can be done in 4C2 = 6 ways.

When you need to have two 1s, pick three and club = 4C3 = 4 ways.

When you need to have three 1s, club all = 1way (ie 1*1*1*210)

And finally without 1s, it is the same as above 2*3*5*7.

Now, if you need to find the number of solutions, for the first case, the 4 distinct factors can be futher arranged in 4! ways.

So 4C2*4!

Similarly, second case, where two ones are there 4C3*(4!/2!)

And third Case 1*(4!/3!)

And fourth case again 4! (all are distinct)

Add all these cases, and you will get the number of solutions as: 24*6 + 4*12 + 4 + 24 = 220 solutions.

Hope that helps.

Best Regards,

Ashwin (IIT Madras).

Hi Basit,

My Previous solution had an error in one particular case (Pls ignore)....

First factorise 210 = 2*3*5*7.

Now, this has no "1".

When you want to have one "1". Then from the above 4 factors, pick two and club it. Which can be done in 4C2 = 6 ways. (an example would be 1*6*5*7)

When you need to have two 1s, pick three factors and club = 4C3 = 4 ways, or club any two factors and the remaining two factors into another = 4C2 ways. An Eg (ie 1*1*30*7 or 1*1*6*35)

When you need to have three 1s, club all = 1way (ie 1*1*1*210)

And finally without 1s, it is the same as the original factors 2*3*5*7.

Now, if you need to find the number of solutions, for the first case, the 4 distinct factors can be futher arranged in 4! ways.

So 4C2*4!

Similarly, second case, where two ones are there (4C3+4C2)*(4!/2!)

And third Case 1*(4!/3!)

And fourth case again 4! (all are distinct)

Add all these cases, and you will get the number of solutions as: 24*6 + 10*12 + 4 + 24 = 292 solutions (if order of x1,x2,x3,x4 is important....)

If order is not imoprtant: (It would be 6+10+1+1 = 18 solutions)

Hope that helps.

Best Regards,

Ashwin (IIT Madras).

Wats the ans,is it 256?