What's the coefficient of the term?

Question

Hi everyone I have a problem, I can't find the coefficient of $c_{1}^{2}c_{2}^{2}c_{3}^{2}$ in this expression $(c_{1}+c_{2}+c_{3})^{2}(c_{1}^{2}+c_{2}^{2}+c_{3}^{2})^{2}$

It's in a discrete maths exercise if this helps. It seems like Newton's binomial but I can't find a solution

Answer 1

We have

$$(c_1+c_2+c_3)^2=c_1^2+c_2^2+c_3^2+2c_1c_2+2c_1c_3+2c_2c_3$$

and similarly, we have

$$(c_{1}^{2}+c_{2}^{2}+c_{3}^{2})^{2}=c_1^4+c_2^4+c_3^4+2c_1^2c_2^2+2c_1^2c_3^2+2c_2^2c_3^2$$

Now let's think about what we get from multiplying both equations. The $c_1^2$ in the first equation goes with $2c_2^2c_3^2$ to give $2c_1^2c_2^2c_3^2$. Similarly for $c_2^2$ and for $c_3^2$, so that we already have $6c_1^2c_2^2c_3^2$.

Now about the other terms of the first equation, it is rather clear that we can not get $c_1^2c_2^2c_3^2$ out of them (as multiplying them with one of the term ofthe second equation will lie an odd exponent for at least one of the $c_i$).

Hence, the coefficient you are looking for is just $6c_1^2c_2^2c_3^2$, as there are no other possibilities.

Answer 2

Note that the right factor can only generate terms with squares, so we need only consider the square terms generated by the left factor.

Hence the terms involving $c_1^2,c_2^2,c_3^2$ simultaneously are

$$c_1^2\,2c_2^2c_3^2+c_2^2\,2c_3^2c_1^2+c_3^2\,2c_1^2c_2^2.$$

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For your curiosity,

$$c_1^6+4c_1^3c_2^3+4c_1^3c_3^3+3c_1^2c_2^4+6c_1^2c_2^2c_3^2+3c_1^2c_3^4+2c_1c_2^5+4c_1c_2^2c_3^3+2c_1c_2c_3^4+2c_1c_3^5+4c_1c_3^2c_2^3+2c_1c_3c_2^4+c_2^6+4c_2^3c_3^3+3c_2^2c_1^4+3c_2^2c_3^4+2c_2c_1^5+4c_2c_1^2c_3^3+2c_2c_3^5+4c_2c_3^2c_1^3+2c_2c_3c_1^4+c_3^6+3c_3^2c_1^4+3c_3^2c_2^4+2c_3c_1^5+4c_3c_1^2c_2^3+2c_3c_2^5+4c_3c_2^2c_1^3$$