Party problems coming soon!

Consider the Pascal's triangle in the figure where the binomial coefficients are arranged in the usual manner. Select any binomial coefficient from anywhere except the right edge of the triangle and labet it $C$ . To the right of $C$ , in the horizontal line, there are $t$ numbers, we denote them as $a_1,a_2,\cdots,a_t$ , where $a_t = 1$ is the last number of the series. Consider the line parallel to the left edge of the triangle containing $C$ , there will only be $t$ numbers diagonally above $C$ in that line. We successively name them as $b_1,b_2,\cdots,b_t$ , where $b_t = 1$ . Find the value of

$b_ta_1-b_{t-1}a_2+b_{t-2}a_3-\cdots+(-1)^{t-1}b_1a_t$ .

For example, Suppose you choose $\binom41 = 4$ (see figure), then $t = 3$ , $a_1 = 6, a_2 = 4, a_3 = 1$ and $b_1 = 3, b_2 = 2, b_3 = 1$ . $\begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1 & & \underset{b_3}{1} & & & & \\ & & & 1 & & \underset{b_2}{2} & & 1 & & & \\ & & 1 & & \underset{b_1}{3} & & 3 & & 1 & & \\ & 1 & & \boxed{4} & & \underset{a_1}{6} & & \underset{a_2}{4} & & \underset{a_3}{1} & \\ \ldots & & \ldots & & \ldots & & \ldots & & \ldots & & \ldots \\ \end{array}$

This problem is original.