No. of positive integral solutions of

              x1+x2+x3+x4+x5 = x1*x2*x3*x4*x5   where * stands for multiplication

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iam a new user ,i wish to get some intresting questions in matrics.....pls help me..

No Rudra it is not like that

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I think the answer is (r-1)(r-2)(r-3)(r-4)/24 where r=x₁x₂x₃x₄x₅.Is it correct?

wat is 'r'?

look me really not getting ne way out of dis one....but neways ill post wat i was trying....

let us concider a 5rd degree eqn. of  x⁵+ax⁴+bx³+cx²+dx+e.....having integral roots....say x1 x2 x3...x5....

E == summation (formula edittor not working)

den E (x1) = -a......E (x1.x2) =b......E (x1.x2.x3)=-c.....E (x1.x2.x3.x4)=d.....x1.x2.x3.x4.x5=-e

we have -e=-a or e=a.....nd we also have { E (x1.x2.x3) / (-e)} = E (1/(x1.x2)) =c/e

using am >= hm for x1.x2,  x2.x3, .......

we have the result b/(⁵C₂) >= (⁵C₂)/(c/e)--------------1)

nd going in same lines we also have E (x1.x2)/ (-e) = E (1/x1x2x3) = (-b/e)

and applying am >= hm for x1.x2.x3, x2.x3.x4,........

we have -c/(⁵C₃) >= (⁵C₃)/(-b/e)--------------------------2)

using dese 2 eqn.s and also the fact dat a=e.....we cud probably get sumwhere....

sorry....dis im not able to get beyond dis...

i may be totally wrong...

please help me out...