Coefficient conundrum

We have to find coefficient of $a^{8}b^{4}c^{9}d^{9}$ in $(abc+abd+acd+bcd)^{10}$ .

It is same as if we have to find the coefficient of $\dfrac{a^{10}b^{10}c^{10}d^{10}}{a^{2}b^{6}cd}$ in $a^{10}b^{10}c^{10}d^{10}\Big( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\Big)^{10}$ .

It is as same as if we have to find the coefficient of $\dfrac{1}{a^{2}b^{6}cd}$ in $\Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\Big)^{10}$ .

Now we know that coefficient is $\binom{10}{6} \binom{4}{2} \binom{2}{1} = 2520.$ .

So the sum of digits of $2520$ is $2+5+2 + 0 = 9$ .

We have a⁸b⁴c⁹d⁹. Now we must get this in terms of abc, abd, acd, bcd Hence, that will be= (abc)¹(abd)¹(acd)⁶(bcd)² Now applying multinomial theorum we can get (abc+abd+acd+bcd)¹⁰= Σ(₁‚₁,¹⁰₆,₂)(abc)¹(abd)¹(acd)⁶(bcd)² Hence, coefficient of (abc)¹(abd)¹(acd)⁶(bcd)² is Σ(₁‚₁,¹⁰₆,₂)= 10!/(1! 1! 6!*2!)=2520 Coefficient= 2520, sum of 2520= 2+5+2+0= 9 Answer=9

You must first realise that each of the terms is just all of the terms excluding 1 variable (which is different for each term), so if we set up 10 boxes with a copy of those terms each where we have to select 2 term from each box to multiply, selecting n of 1 term will mean we have to have to select 10-n copies for other terms, meaning that whatever variable is secluded from that box we are selecting n from must be raised to 10-n in our expanded term as all other boxes have that term, so we can quickly see that there is only 1 viable selection (we have to select 10-k of the term where k is the value by which the variable which is excluded in the term is raised by in our expanded term, for all terms) to get our, which term is, in the order of the expression in the question, 1 1 6 2, and by the multinational theorem we hence know the coefficient is 10!/(1! 1! 6! 2!)=10 9 8 7/2=10 9 56/2=10*504/2=5040/2=2520, 2+5+2+0=9