A) \[\frac{x+y!}{r!}\]
B) \[^{x+y}{{C}_{r}}^{xy}{{C}_{r}}\]
C) \[^{x+y}{{C}_{r}}\]
D) \[^{xy}{{C}_{r}}\]
Correct Answer: C
Solution :
\[{{(1+z)}^{x}}=1{{+}^{x}}{{C}_{1}}z{{+}^{x}}{{C}_{2}}{{z}^{2}}{{+}^{x}}{{C}_{3}}{{z}^{3}}\] \[+....{{+}^{x}}{{C}_{r}}{{z}^{r}}+....\] ?(i) \[{{(1+z)}^{y}}=1{{+}^{y}}{{C}_{1}}z{{+}^{y}}{{C}_{2}}{{z}^{2}}{{+}^{y}}{{C}_{3}}{{z}^{3}}+...\] \[{{+}^{y}}{{C}_{r}}{{z}^{r}}+...\] ?.(ii) \[{{(1+z)}^{x+y}}=1{{+}^{x+y}}{{C}_{1}}z{{+}^{x+y}}{{C}_{2}}{{z}^{2}}+...\] \[^{x+y}{{C}_{3}}{{z}^{3}}+....{{+}^{x+y}}{{C}_{r}}{{z}^{r}}....\] ...(iii) Coefficient of\[{{z}^{r}}\]in the multiplication of Eqs. (i) and (ii), \[^{x}{{C}_{r}}{{+}^{x}}{{C}_{r-1}}^{y}{{C}_{1}}{{+}^{x}}{{C}_{r-2}}^{y}{{C}_{2}}+....{{+}^{y}}{{C}_{r}}\] Hence, from Eq. (iii), \[^{x+y}{{C}_{r}}{{=}^{x}}{{C}_{r}}{{+}^{x}}{{C}_{r-1}}^{y}{{C}_{1}}+....{{+}^{y}}{{C}_{r}}\]
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