Multinomial Coefficient

Using the multinomial theorem:
$$(1 + x + x^2)^{10} = \sum_{i+j+k=10} \binom{10}{i,j,k} 1^i \cdot x^j \cdot (x^2)^k = \sum_{i+j+k=10} \binom{10}{i,j,k} x^{j+2k}$$

For the coefficient of $$x^5$$, we need: $$j + 2k = 5$$ and $$i + j + k = 10$$

From the first equation: $$j = 5 - 2k$$
From the second equation: $$i = 10 - j - k = 10 - (5-2k) - k = 5 + k$$

For non-negative integers:
- $$j = 5 - 2k \geq 0 \Rightarrow k \leq 2.5 \Rightarrow k \leq 2$$
- $$k \geq 0$$

So $$k \in \{0, 1, 2\}$$

For $$k = 0$$: $$i = 5, j = 5, k = 0$$
$$\binom{10}{5,5,0} = \frac{10!}{5! \cdot 5! \cdot 0!} = \binom{10}{5} = 252$$

For $$k = 1$$: $$i = 6, j = 3, k = 1$$
$$\binom{10}{6,3,1} = \frac{10!}{6! \cdot 3! \cdot 1!} = 840$$

For $$k = 2$$: $$i = 7, j = 1, k = 2$$
$$\binom{10}{7,1,2} = \frac{10!}{7! \cdot 1! \cdot 2!} = 360$$

Therefore the final coefficient of $$x^5$$ in the expansion of $$(1 + x + x^2)^{10}$$ $$= 360 + 840 + 252=1452$$

To obtain the coefficient of $$x^4$$ from $$(1-x+2x^2)^5$$

let $$a+b+c=5$$ where $$a$$ copies of $$1$$,

$$b$$ copies of $$-x$$, and

$$c$$ copies of $$2x^2$$ are selected.

The power of $$x$$ contributed is $$b+2c$$

Hence we require

$$b+2c=4$$

The possible solutions are:

$$ (b,c)=(4,0) $$

$$ (b,c)=(2,1) $$

$$ (b,c)=(0,2) $$

For

$$ (b,c)=(4,0) $$

we have

$$a=1$$

Contribution

$$=\frac{5!}{1!4!0!}(-1)^4$$

$$=5$$

For

$$ (b,c)=(2,1) $$

we have

$$a=2$$

Contribution

$$=\frac{5!}{2!2!1!}(-1)^2(2)^1$$

$$=30\times2$$

$$=60$$

For

$$ (b,c)=(0,2) $$

we have

$$a=3$$

Contribution

$$=\frac{5!}{3!0!2!}(2)^2$$

$$=10\times4$$

$$=40$$

Therefore the coefficient of

$$x^4$$

is

$$5+60+40$$

$$=105$$

Final Answer :

$$105$$