IIT JEE , AIEEE Forum - Mathematics , Algebra

elastiboysai

K fine'
Im not verry sure..neways,
the approach i had given u earlier was my own, bt its quite long,
heres anoder method
2^3*3*5*7
=x1x2x3x4x5
now3,5,7 can be assigned 2 any of d 5 variables, 5^3
as for 2^3
it can b assigned to 1 variable in 5 ways, 2 variables in 10*2 ways (5c2) *2 for
(1,2) and (2,1) they r diffn here
3 variables in 10 ways (5c3) thus total=35 ways,
now total no of ways=
35*5*5*5=4375
edited:
gokuls rt
i hadnt taken care of (2,1) and (1,2) permutation wen i distr. 2^3 to 2 variables
now its corrected

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