IIT JEE , AIEEE Forum - Mathematics , Algebra

nadeemoidu

It is the same as finding the coefficient of $\frac{(abc)^6}{a^3b^2c}$ in the expansion of $( \frac{1}{a}+\frac{1}{b}+\frac{1}{c})^6$

risin

Could you pl elaborate and proceed further.

nadeemoidu

This is now equivalent to finding the coefficient of $(\frac{1}{a})^3(\frac{1}{b})^2(\frac{1}{c})$ in the expansion of $( \frac{1}{a}+\frac{1}{b}+\frac{1}{c})^6$ . This is the usual multinomial expansion .

The answer is 6! / 3! 2! = 60

risin

Phew..thanks..

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