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

We have to find the coefficient of the term containing $$a^{7} b^{8}$$ in the expansion of $$(a + 2b + 4ab)^{10}$$.

First recall the multinomial theorem. For any three terms $$x_{1}, x_{2}, x_{3}$$,

$$ (x_{1} + x_{2} + x_{3})^{10} = \sum_{n_{1}+n_{2}+n_{3}=10} \dfrac{10!}{n_{1}! \, n_{2}! \, n_{3}!}\; x_{1}^{\,n_{1}} x_{2}^{\,n_{2}} x_{3}^{\,n_{3}}. $$

In our question we identify

$$x_{1}=a, \qquad x_{2}=2b, \qquad x_{3}=4ab.$$

Let $$n_{1}, n_{2}, n_{3}$$ be the number of times these three terms are chosen, so

$$n_{1}+n_{2}+n_{3}=10.$$

The powers of $$a$$ and $$b$$ contributed by one choice of each term are

$$x_{1}=a \;\; \longrightarrow \;\; a^{1} b^{0},$$ $$x_{2}=2b \;\; \longrightarrow \;\; a^{0} b^{1},$$ $$x_{3}=4ab \;\;\longrightarrow \;\; a^{1} b^{1}.$

Hence after making $$n_{1}, n_{2}, n_{3}$$ selections, the total exponents of $$a$$ and $$b$$ become

$$\text{Power of } a : \; n_{1} + n_{3},$$ $$\text{Power of } b : \; n_{2} + n_{3}.$$

We need these powers to be $$7$$ and $$8$$ respectively, so we write the two equations

$$n_{1} + n_{3} = 7, \qquad n_{2} + n_{3} = 8.$$

Together with $$n_{1}+n_{2}+n_{3}=10,$$ we now solve for $$n_{1}, n_{2}, n_{3}.$

Adding the first two equations gives

$$n_{1}+n_{2}+2n_{3}=15.$$

But $$n_{1}+n_{2}=10-n_{3},$$ so substituting,

$$(10-n_{3}) + 2n_{3} = 15 \;\; \Longrightarrow \;\; 10 + n_{3} = 15 \;\; \Longrightarrow \;\; n_{3} = 5.$$

Now we back-substitute:

$$n_{1} = 7 - n_{3} = 7 - 5 = 2,$$ $$n_{2} = 8 - n_{3} = 8 - 5 = 3.$$

Thus the required term arises from $$n_{1}=2,\, n_{2}=3,\, n_{3}=5.$

According to the multinomial theorem, the coefficient of this term is

$$\dfrac{10!}{n_{1}! \, n_{2}! \, n_{3}!}\; (2)^{n_{2}}\;(4)^{n_{3}}.$$

Writing out the factorials,

$$10! = 3628800,$$ $$n_{1}! = 2! = 2,$$ $$n_{2}! = 3! = 6,$$ $$n_{3}! = 5! = 120.$$

So

$$\dfrac{10!}{2!\,3!\,5!} = \dfrac{3628800}{2 \times 6 \times 120} = \dfrac{3628800}{1440} = 2520.$$

Next incorporate the numerical factors from the bases:

$$(2)^{n_{2}} = 2^{3} = 8,$$ $$(4)^{n_{3}} = (2^{2})^{5} = 2^{10} = 1024.$$

Multiplying these powers of $$2$$,

$$8 \times 1024 = 2^{3} \times 2^{10} = 2^{13} = 8192.$$

Hence the full numerical coefficient of the term $$a^{7} b^{8}$$ is

$$2520 \times 8192 = 20643840.$$

We are told in the question that this coefficient can be written as $$K \cdot 2^{16}.$$ Since $$2^{16}=65536,$$ we set up the equation

$$K \times 65536 = 20643840.$$

Dividing to isolate $$K,$$

$$K = \dfrac{20643840}{65536}.$$

Compute the division step by step:

$$65536 \times 300 = 19660800,$$

Subtracting, $$20643840 - 19660800 = 983040,$$

$$65536 \times 10 = 655360,$$

Subtracting, $$983040 - 655360 = 327680,$$

$$65536 \times 5 = 327680,$$

Subtracting, the remainder is $$0.$$

Adding the multipliers, $$300 + 10 + 5 = 315.$$ Therefore,

$$K = 315.$$

So, the answer is $$315$$.