The value of 50C4 +∑r=1656−rC3 is

A
$^{55} C_{4}$
B
$^{55} C_{3}$
C
$^{56} C_{3}$
D
$^{56} C_{4}$

Solution:

$^{50} C_{4} + \sum_{r = 1}^{6}^{56 - r} C_{3}$ = $^{50} c_{4} +^{55} c_{3} +^{54} c_{3} +^{53} c_{3} + - - - +^{50} c_{3}$

$\begin{array}{l} = \{C (50, 4) + C (50, 3)\} + C (51, 3) + C (52, 3) + ......... + C (55, 3) [∵ C (n, r) + C (n, r - 1) = C ((n + 1), r)] \\ = \{C (51, 4) + C (51, 3)\} + C (52, 3) + ...... + C (55, 3) \\ = C (52, 4) + C (52, 3) + ......... + C (55, 3) \\ .................. \\ = C (56, 4) \end{array}$

$=^{56} c_{4}$

The value of $^{50} C_{4} + \sum_{r = 1}^{6}^{56 - r} C_{3}$ is

Solution:

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