If the coefficient of a^7b^8 in the expansion of (a+2b+4ab)^10 is k2^16, then k is equal t…

Question

If the coefficient of $a^7b^8$ in the expansion of $(a+2b+4ab)^{10}$ is $k\cdot2^{16}$, then $k$ is equal to

Answer: 315

Accepted Answer

$$(a+2b+4ab)^{10}=a^{10}b^{10}\left(\frac1b+\frac2a+4\right)^{10}$$

General term: $\dfrac{10!}{r_1!r_2!r_3!}\left(\dfrac1b\right)^{r_1}\left(\dfrac2a\right)^{r_2}4^{r_3}$, $r_1+r_2+r_3=10$. For $a^7b^8$: need $a^{10-r_2}=a^7\Rightarrow r_2=3$, $b^{10-r_1}=b^8\Rightarrow r_1=2$, so $r_3=5$.

$$\text{Coeff}=\frac{10!}{2!3!5!}\times2^3\times4^5=\frac{10!}{2!3!5!}\times2^3\times2^{10}=\frac{10!}{2!3!5!}\times2^{13}$$

$$\frac{10!}{2!3!5!}=2520$$... rechecking with the given result $k\cdot2^{16}$: $\dfrac{10!}{2!3!5!}\times2^{13}=2520\times2^{13}$; but $2520=315\times8=315\times2^3$, so total $=315\times2^{16}$.

$k=315$

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