Binomial Theorem

We will expand $(1+x-x^2)^6$ and find the coefficients of $x^4$ and $x^6$.Multimonial expansion states that$\;(a_1+a_2+a_3)^n=\sum\;\frac{n!}{k_{1}!k_{2}!k_{3}!}$$\;a_1^k a_2^{k_2} a_3^{k_3}$where $k_1+k_2+k_3=n$In our case, $a_1=1, a_2=x, a_3=-x^2$ and $n=6$From $k_1+k_2+k_3=6$$\Rightarrow\;\; k_1=6-k_2-k_3$To get $x^4$, we can have different combinations of $x$ and $x^2$$k_1$ : number of times 1 is chosen$k_2$ : number of times $x$ is chosen$k_3$ : number of times $x^2$ is chosenThe equation is: $k_2+2k_3=4$Now, if $k_3=0$, then $k_2=4$ and $k_1=2$If $k_3=1$, then $k_2=2$ and $k_1=3$If $k_3=2$, then, $k_2=0, k_1=4$Now, find the coefficients for each valid combination.Coefficient of$\;x^4=\;\frac{6!}{2!4!0!}-\;\frac{6!}{3!2!1!}+\;\frac{6!}{4!0!2!}$$\;=15-60+15=-30$Now, for $x^6, k_2+2k_3=6$ and$k_1+k_2+k_3=6$If $k_3=0$, then $k_2=6$, and $k_1=0$If $k_3=1$, then $k_2=4$, and $k_1=1$If $k_3=2$ then $k_2=2$ and $k_1=2$If $k_3=3$, then $k_2=0$, and $k_1=3$So, coefficient of $x^6=\;\frac{6!}{0!6!0!}-\;\frac{6!}{1!4!1!}+\;\frac{6!}{2!2!2!}-\;\frac{6!}{3!0!3!}$$=1-30+90-20=41$∴ Sum of the coefficient of $x^4$ and $x^6=41-30=11$