#### Prove by binomial expansion that  $\small { }^m \mathrm{C}_{\mathrm{r}}{ }^n \mathrm{C}_0+{ }^{\mathrm{m}} \mathrm{C}_{\mathrm{r}-1 \cdot} \cdot{ }^n \mathrm{C}_1+{ }^{\mathrm{m}} \mathrm{C}_{\mathrm{r}-2} \cdot{ }^n \mathrm{C}_2+\ldots \ldots+{ }^{\mathrm{m}} \mathrm{C}_0 \cdot{ }^n \mathrm{C}_{\mathrm{r}}=\mathrm{m}+{ }^n \mathrm{C}_{\mathrm{r}}$where r < m or n.    Option: 1 1  Option: 2 ${ }^{m} C_r$Option: 3 ${ }^{m+n} C_r$Option: 4 None of these

$\\ \text { We have }(1+x)^m={ }^m C_0+{ }^m C_1 x+{ }^m C_2 x^2+\ldots \ldots .+{ }^m C_r x^r+\ldots \ldots .+{ }^m C_m x^m$

$(1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+\ldots \ldots+{ }^n C_r x^r+\ldots \ldots .+{ }^n C_n x^n$Multiplying and equating coefficient of $x^r$ on both sides, we get,

$={ }^m C_r \cdot{ }^n C_0+{ }^m C_{r-1} \cdot{ }^n C_1+\ldots \ldots+{ }^m C_0 \cdot{ }^n C_r$

\begin{aligned} & \therefore{ }^m C_r \cdot{ }^n C_0+{ }^m C_{r-1} \cdot{ }^n C_1+\ldots . .+{ }^m C_0 \cdot{ }^n C_r=\text { coefficient of } x^r \text { in }(1+x)^{m+n} \\ & ={ }^{m+n} C_r \end{aligned}