The value of 50C4+∑r=16 56−rC3 is

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

Solution:

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

$\begin{array}{l} =^{50} C_{4} + (^{55} C_{3} +^{54} C_{3} + \dots +^{50} C_{3}) \\ = (^{50} C_{4} +^{50} C_{3}) + (^{51} C_{3} + \dots +^{55} C_{3}) \\ = (^{51} C_{4} +^{51} C_{3}) + (^{52} C_{3} + \dots +^{55} C_{3}) \\ = (^{52} C_{4} +^{52} C_{3}) + (^{53} C_{3} + \dots +^{55} C_{3}) \\ = \dots . . \\ =^{56} C_{4} \end{array}$

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

Solution:

Related content