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  • Prove by binomial expansion that  <img alt="\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}

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


Answers (1)

best_answer

\\ \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^nMultiplying 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}

Posted by

Rakesh

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