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Binomial Theorem Part 1

Section: Mathematics

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Question : 89 of 100

Marks: +1, -0

The value of

{}^{50}C_4 + \sum\limits_{r=1}^{6} {}^{56-r}C_3

${}^{55}C_4$
${}^{55}C_3$
${}^{56}C_3$
${}^{56}C_4$

Solution:

Given,

{}^{50}C_4 + \sum\limits_{n=1}^{6} {}^{56-r}C_3

\Rightarrow {}^{50}C_4 + {}^{55}C_3 + {}^{54}C_3 + {}^{53}C_3 + {}^{52}C_3 + {}^{51}C_3 + {}^{50}C_3

Arrange those this way

\Rightarrow {}^{50}C_4 + {}^{50}C_3 + {}^{51}C_3 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3

We know this formula

\left[ {}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r \right]

which is used to solve this problem.

\Rightarrow {}^{51}C_4 + {}^{51}C_3 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3

\Rightarrow {}^{52}C_4 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3

\Rightarrow {}^{53}C_4 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3

\Rightarrow {}^{54}C_4 + {}^{54}C_3 + {}^{55}C_3

\Rightarrow {}^{55}C_4 + {}^{55}C_3

\Rightarrow {}^{56}C_4

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