#9 Measure Your Calibre

The given expression can be written as

${(bcd+acd+abd+abc)}^7 = {(abcd)}^7 {\left( \dfrac 1a + \dfrac 1b + \dfrac 1c + \dfrac 1d \right)}^7$

Outside the expansion we have the term $a^7 b^7 c^7 d^7$ , thus we have to find the coefficient of the term $\dfrac{1}{a^2 b^2 c^2 d}$ inside the expansion.

By general multinomial theorem, the expansion can be written as

$\displaystyle \sum_{0 \le \alpha,\beta,\gamma,\delta \le 7}^{\alpha+\beta+\gamma+\delta = 7} \dfrac{7!}{\alpha! \beta! \gamma! \delta!} {\left( \dfrac 1a \right)}^{\alpha} {\left( \dfrac 1b \right)}^{\beta} {\left( \dfrac 1c\right)}^{\gamma} {\left( \dfrac 1d\right)}^{\delta}$

Here $\alpha=\beta=\gamma=2$ and $\delta =1$ .

Thus the required coefficient is $\dfrac{7!}{{(2!)}^3} = 630$ .