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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).
My Previous solution had an error in one particular case (Pls ignore)....
First factorise 210 = 2*3*5*7.
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)
And finally without 1s, it is the same as the original factors 2*3*5*7.
Similarly, second case, where two ones are there (4C3+4C2)*(4!/2!)
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)
Wats the ans,is it 256?
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