Show that the number of sequences of length 40 are there using the alphabet {a,b,c,d} Show that the number of sequences of length 40 are there using the alphabet {a,b,c,d} such that number of a's in the sequence is divisible by 3 is $\frac{4^{40}+2Re((3+w)^{40})}{3}$
sequence with 3 a's=$40C3\times3^{(40-3)}$
sequence with 6 a's=$40C6\times 3^{(40-6)}$
sequence with 9 a's=$40C9\times3^{(40-9)}$
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sequence with 36 a's=$40C36\times3^{(40-36)}$
sequence with 39 a's=$40C39\times3^{(40-39)}$
I'm stuck as to where the Re and $w$ come into play
 A: Consider the polynomial $(a+b+c+d)^{40}$. The coefficient of a term of the form $a^{k_1}b^{k_2}c^{k_3}d^{k_4}$ counts the number of sequences of length 40 composed of ${k_1}$ $a$'s, ${k_2}$ $b$'s, ${k_3}$ $c$'s, ${k_2}$ $d$'s. Therefore we are required to find the sum of coefficients of the terms of $(a+b+c+d)^{40}$ with the exponent of a divisible by the $3$.
To do this, we set $b=c=d=1$ in the polynomial as there is no restriction on the number of $b$'s, $c$'s or $d$'s. Thus our answer is the sum of coefficients of the terms with exponent divisible by $3$ of the polynomial $(a+3)^{40}$.
Let $f(a)=(a+3)^{40}=\sum_{i=0}^{40}{l_i}a^{i}$. Therefore we have to get hold of $z=l_0+l_3+l_6+...+l_{39}$. Notice $f(1)= l_0+l_1+l_2+...+l_{40}$, $f(\omega)=l_0+l_1\omega+l_2\omega^{2}+l_3+l_4\omega+...+l_{40}\omega$. $f(\omega^2)=l_0+l_1\omega^{2}+l_2\omega+l_3+l_4\omega^{2}+...+l_{40}\omega^{2}$.
Therefore $f(1)+f(\omega)+f(\omega^{2})=3(l_0+l_3+l_6+...+l_{39})=3z$
Hence, $$z=\frac{4^{40}+(3+\omega)^{40}+(3+\omega^{2})^{40}}{3}
=\frac{4^{40}+(3+\omega)^{40}+(3+\bar \omega)^{40}}{3}
=\frac{4^{40}+2Re((3+\omega)^{40})}{3}$$
Q.E.D.
