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Find the value of

\(^{50}C_4 + \displaystyle\sum_{r=1}^{6} \space^{56-r}C_3\)

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Best answer

Given expression is

\(^{50}C_4 + \displaystyle\sum_{r=1}^{6} \space^{56-r}C_3\)

\(=\space^{56}C_4+^{55}C_3+^{54}C_3+^{53}C_3+^{52}C_3+^{51}C_3+^{50}C_3\)

Writing the terms in reverse order, we get 

=\((^{50}C_4 +^{50}C_3)+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3\)= \(^{50}C_4+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3\)

[∴ \(^nC_r+^nC_{r-1}=^{n+1}C_r\)

= \((^{51}C_4+^{51}C_3)+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3\)

= \((^{52}C_4+^{52}C_3)+^{53}C_3+^{54}C_3+^{55}C_3\)

= \((^{53}C_4+^{53}C_3)+^{54}C_3+^{55}C_3\)

= \((^{54}C_4+^{54}C_3)+^{55}C_3\)

= \(^{55}C_4+^{55}C_3\)

= \(^{56}C_4\)

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