Write the value of

Question:

Write the value of $\sum_{r=1}^{6}{ }^{56-r} C_{3}+{ }^{50} C_{4}$.

Solution:

We know:

nCr-">-1 + nCr = n+1Cr

Now, we have:

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

$={ }^{55} C_{3}+{ }^{54} C_{3}+{ }^{53} C_{3}+{ }^{52} C_{3}+{ }^{51} C_{3}+{ }^{50} C_{3}+{ }^{50} C_{4}$

$={ }^{55} C_{3}+{ }^{54} C_{3}+{ }^{53} C_{3}+{ }^{52} C_{3}+{ }^{51} C_{3}+{ }^{51} C_{4}$

$={ }^{55} C_{3}+{ }^{54} C_{3}+{ }^{53} C_{3}+{ }^{52} C_{3}+{ }^{52} C_{4}$

$={ }^{55} C_{3}+{ }^{54} C_{3}+{ }^{53} C_{3}+{ }^{53} C_{4}$

$={ }^{55} C_{3}+{ }^{54} C_{3}+{ }^{54} C_{4}$

$={ }^{55} C_{3}+{ }^{55} C_{4}$

$={ }^{56} C_{4}$

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