The value of $\sum_{r=0}^{20}{ }^{50-r} C_{6}$ is equal to:
Correct Option: 1
The given series, $\sum_{r=0}^{20}{ }^{50-r} C_{6}$
$={ }^{50} C_{6}+{ }^{49} C_{6}+{ }^{48} C_{6}+{ }^{47} C_{6}+\ldots+{ }^{32} C_{6}+{ }^{31} C_{6}+{ }^{30} C_{6}$
$=\left({ }^{30} C_{7}+{ }^{30} C_{6}\right)+{ }^{31} C_{6}+{ }^{32} C_{6}+\ldots .+{ }^{49} C_{6}+{ }^{50} C_{6}-{ }^{30} C_{7}$
$=\left({ }^{31} C_{7}+{ }^{31} C_{6}\right)+{ }^{32} C_{6}+\ldots+{ }^{49} C_{6}+{ }^{50} C_{6}-{ }^{30} C_{7}$
$=\left({ }^{32} C_{7}+{ }^{32} C_{6}\right)+\ldots .+{ }^{49} C_{6}+{ }^{50} C_{6}-{ }^{30} C_{7}$
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$={ }^{51} C_{7}-{ }^{30} C_{7}$