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If ${ }^{n-3} C_r+B^{n-3} C_{r-1}+B^{\prime n-3} C_{r-2}+{ }^{n-3} C_{r-3}$ $={ }^n C_r$ holds. for all $n \geq r \geq 3$, then $\left(B \cdot B^{\prime}\right)=$.
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The correct answer is:
$(3,3)$
${ }^{n-3} C_r+B^{n-3} C_{r-1}+B^{\prime n-3} C_{r-2}$
$+{ }^{n-3} C_{r-3}={ }^n C_r$
$\Rightarrow\left({ }^{n-3} C_r+{ }^{n-3} C_{r-1}\right)+\left({ }^{n-3} C_{r-1}+{ }^{n-3} C_{r-2}\right)$ $+\left({ }^{n-3} C_{r-2}+{ }^{n-3} C_{r-3}\right)$ $+(B-2){ }^{n-3} C_{r-1}+\left(B^{\prime}-2\right){ }^{n-3} C_{r-2}={ }^n C_3$
$\Rightarrow\left[{ }^{n-2} C_r+{ }^{n-2} C_{r-1}\right]+{ }^{n-2} C_{r-2}$ $+\left({ }^{n-3} C_{r-1}+{ }^{n-3} C_{r-2}\right)+(B-3){ }^{n-3} C_{r-1}$ $+\left(B^{\prime}-3\right){ }^{n-3} C_{r-2}={ }^n C_r$
$\Rightarrow{ }^{n-1} C_r+\left[{ }^{n-2} C_{r-2}+{ }^{n-2} C_{r-1}\right]$ $+(B-3)^{n-3} C_{r-2}+\left(B^{\prime}-3\right){ }^{n-3} C_{r-2}={ }^n C_r$
$\begin{aligned} & \Rightarrow{ }^n C_r+(B-3){ }^{n-3} C_{r-1}+\left(B^{\prime}-3\right)^{n-3} C_{r-2}={ }^n C_r \\ & \Rightarrow(B-3){ }^{n-3} C_{r-1}+\left(B^{\prime}-3\right){ }^{n-3} C_{r-2}=0\end{aligned}$
It is only possible when, $B-3=0$ and $B^{\prime}-3=0$ $\therefore\left(B, B^{\prime}\right) \equiv(3,3)$
$+{ }^{n-3} C_{r-3}={ }^n C_r$
$\Rightarrow\left({ }^{n-3} C_r+{ }^{n-3} C_{r-1}\right)+\left({ }^{n-3} C_{r-1}+{ }^{n-3} C_{r-2}\right)$ $+\left({ }^{n-3} C_{r-2}+{ }^{n-3} C_{r-3}\right)$ $+(B-2){ }^{n-3} C_{r-1}+\left(B^{\prime}-2\right){ }^{n-3} C_{r-2}={ }^n C_3$
$\Rightarrow\left[{ }^{n-2} C_r+{ }^{n-2} C_{r-1}\right]+{ }^{n-2} C_{r-2}$ $+\left({ }^{n-3} C_{r-1}+{ }^{n-3} C_{r-2}\right)+(B-3){ }^{n-3} C_{r-1}$ $+\left(B^{\prime}-3\right){ }^{n-3} C_{r-2}={ }^n C_r$
$\Rightarrow{ }^{n-1} C_r+\left[{ }^{n-2} C_{r-2}+{ }^{n-2} C_{r-1}\right]$ $+(B-3)^{n-3} C_{r-2}+\left(B^{\prime}-3\right){ }^{n-3} C_{r-2}={ }^n C_r$
$\begin{aligned} & \Rightarrow{ }^n C_r+(B-3){ }^{n-3} C_{r-1}+\left(B^{\prime}-3\right)^{n-3} C_{r-2}={ }^n C_r \\ & \Rightarrow(B-3){ }^{n-3} C_{r-1}+\left(B^{\prime}-3\right){ }^{n-3} C_{r-2}=0\end{aligned}$
It is only possible when, $B-3=0$ and $B^{\prime}-3=0$ $\therefore\left(B, B^{\prime}\right) \equiv(3,3)$
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