Co-eff. of $${x^{11}}$$ in exp. of $${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$$
$$ = \left( {{\text{co-eff}}{\text{. of }}{x^a}} \right) \times \left( {{\text{co-eff}}{\text{. of }}{x^b}} \right) \times \left( {{\text{co-eff}}{\text{. of }}{x^c}} \right)$$
Such that $$a + b + c = 11$$
Here $$a = 2m, b = 3n, c = 4p$$
∴ $$2m + 3n + 4p = 11$$
Case l : $$m = 0, n = 1, p = 2$$
Case ll : $$m = 1, n = 3, p = 0$$
Case III : $$m = 2, n = 1, p = 1$$
Case IV : $$m = 4, n = 1, p = 0$$
∴ Required co-eff.
$$\eqalign{ & = {\,^4}{C_0} \times {\,^7}{C_1} \times {\,^{12}}{C_2} + {\,^4}{C_1} \times {\,^7}{C_3} \times {\,^{12}}{C_0} + {\,^4}{C_2} \times {\,^7}{C_1} \times {\,^{12}}{C_1} + {\,^4}{C_4} \times {\,^7}{C_1} \times {\,^{12}}{C_0} \cr & = 462 + 140 + 504 + 7 \cr & = 1113 \cr} $$

$${\left( {a - b} \right)^n},n \geqslant 5$$
In binomial expansion of above $${T_5} + {T_6} = 0$$
$$\eqalign{ & \Rightarrow \,{\,^n}{C_4}{a^{n - 4}}{b^4} + {\,^n}{C_5}{a^{n - 5}}{b^5} = 0 \cr & \Rightarrow \,\,\frac{{^n{C_4}}}{{^n{C_5}}}.\frac{a}{b} = 1 \cr & \Rightarrow \,\,\frac{{4 + 1}}{{n - 4}}.\frac{a}{b} = 1 \cr & \Rightarrow \,\,\frac{a}{b} = \frac{{n - 4}}{5} \cr} $$

$${\left( {1 + x} \right)^j} = 1 + {\,^j}{C_1}x + {\,^j}{C_2}{x^2} + {\,^j}{C_3}{x^3} + ..... + {\,^j}{C_{100}}{x^{100}} + ..... + {\,^j}{C_{200}}{x^{200}}$$
$$\therefore $$ Coefficient of $$x^{100}$$ in the expansion of $${\left( {1 + x} \right)^j} = {\,^j}{C_{100}}$$
Coefficient of $$x^{100}$$ in the expansion of
$$\sum\limits_{j = 0}^{200} {{{\left( {1 + x} \right)}^j}} $$   will be equal to $$\sum\limits_{j = 100}^{200} {^j{C_{100}}} $$
$$ = {\,^{100}}{C_{100}} + {\,^{101}}{C_{100}} + {\,^{102}}{C_{100}} + ..... + {\,^{200}}{C_{100}}$$
\[ = {\,^{200}}{C_{100}} = \left( {\begin{array}{*{20}{c}} {200}\\ {100} \end{array}} \right)\]

The given series is
$$\eqalign{ & 1 + \frac{{1 + a}}{{2!}} + \frac{{1 + a + {a^2}}}{{3!}} + \frac{{1 + a + {a^2} + {a^3}}}{{4!}} + .....\, \cr & {\text{Here, }}{T_n} = \frac{{1 + a + {a^2} + {a^3} + .....\,\,{\text{to }}n{\text{ terms}}}}{{n!}} \cr & = \frac{{1\left( {1 - {a^n}} \right)}}{{\left( {1 - a} \right)\left( {n!} \right)}} = \frac{1}{{1 - a}}\left( {\frac{{1 - {a^n}}}{{n!}}} \right) \cr & \therefore {T_1} + {T_2} + {T_3} + .....{\text{ to }}\infty \cr & = \frac{1}{{1 - a}}\left[ {\frac{{1 - a}}{{1!}} + \frac{{1 - {a^2}}}{{2!}} + \frac{{1 - {a^3}}}{{3!}} + .....{\text{ to }}\infty } \right] \cr & = \frac{1}{{1 - a}}\left[ {\left( {\frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + .....{\text{ to }}\infty } \right) - \left( {\frac{a}{{1!}} + \frac{{{a^2}}}{{2!}} + \frac{{{a^3}}}{{3!}} + .....{\text{ to }}\infty } \right)} \right] \cr & = \frac{1}{{1 - a}}\left[ {\left( {e - 1} \right) - \left( {{e^a} - 1} \right)} \right] = \frac{{e - {e^a}}}{{1 - a}} = \frac{{{e^a} - e}}{{a - 1}} \cr} $$

$$\eqalign{ & {S_2} = \sum\limits_{j = 1}^{10} {j\,{\,^{10}}{C_j} = \sum\limits_{j = 1}^{10} {10} \,{\,^9}{C_{j - 1}}} \cr & = 10\left[ {^9{C_0} + {\,^9}{C_1} + {\,^9}{C_2} + ..... + {\,^9}{C_9}} \right] = 10.\,{2^9} \cr} $$

$$\eqalign{ & {t_{r + 2}} = {\,^{2n}}{C_{r + 1}}{x^{r + 1}};{t_{3r}} = {\,^{2n}}{C_{3r - 1}}{x^{3r - 1}} \cr & {\text{Given}}{{\text{ }}^{2n}}{C_{r + 1}} = {\,^{2n}}{C_{3r - 1}}; \cr & \Rightarrow \,{\,^{2n}}{C_{2n - \left( {r + 1} \right)}} = {\,^{2n}}{C_{3r - 1}} \cr & \Rightarrow \,\,2n - r - 1 = 3r - 1 \cr & \Rightarrow \,\,2n = 4r \cr & \Rightarrow \,\,n = 2r \cr} $$

$$\eqalign{ & {T_{r + 1}} = {\,^{256}}{C_r}{\left( {\sqrt 3 } \right)^{256 - r}}{\left( {^8\sqrt 5 } \right)^r} \cr & \,\,\,\,\,\,\, = {\,^{256}}{C_r}{\left( 3 \right)^{\frac{{256 - r}}{2}}}{\left( 5 \right)^{\frac{r}{8}}} \cr} $$
Terms will be integral if $$\frac{{256 - r}}{2}\& \frac{r}{8}$$   both are +ve integer, which is so if $$r$$ is an integral multiple of 8. As $$0 \leqslant r \leqslant 256$$
∴ $$r$$ = 0, 8, 16, 24, . . . . . , 256, total 33 values.

$$\eqalign{ & {\left( {1 + x + y - z} \right)^9} = {\left\{ {\left( {1 - z} \right) + \left( {x + y} \right)} \right\}^9} \cr & {\left( {1 + x + y - z} \right)^9} = {\,^9}{C_0}{\left( {1 - z} \right)^9} + {\,^9}{C_1}{\left( {1 - z} \right)^8}\left( {x + y} \right) + {\,^9}{C_2}{\left( {1 - z} \right)^7}{\left( {x + y} \right)^2} + ..... + {\,^9}{C_7}{\left( {1 - z} \right)^2} \cdot {\left( {x + y} \right)^7} + .....\,\,. \cr} $$
$${x^3}{y^4}$$  appears in $${\left( {x + y} \right)^7}$$  only.
∴ the required co-efficient $$ = {\,^9}{C_7}\left( { - 2} \right) \cdot {\,^7}{C_4}.$$

We have $${t_{p + 1}} = {\,^{p + q}}{C_p}{x^p}\,{\text{and }}{t_{q + 1}} = {\,^{p + q}}{C_q}{x^q}$$
$$^{p + q}{C_p} = {\,^{p + q}}{C_q}.\left[ {{\text{Remember}}{{\text{ }}^n}{C_r} = {\,^n}{C_{n - r}}} \right]$$

$$r \cdot {\,^n}{C_r} = n \cdot {\,^{n - 1}}{C_{r - 1}}.$$
∴ the sum $$ = n\left\{ {^{n - 1}{C_0} + {\,^{n - 1}}{C_1} + {\,^{n - 1}}{C_2} + ..... + {\,^{n - 1}}{C_{n - 1}}} \right\}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = n \cdot {2^{n - 1}}.$$