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$^{50}C_4+\sum\limits^{6}_{r=1}$$ ^{56-r}C_3=?$

$\begin{array}{1 1}(A) ^{55}C_3 \\(B) ^{55}C_4 \\ (C) ^{56}C_3 \\ (D) ^{56}C_4 \end{array}$

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  • $^nC_r+^nC_{r-1}=^{n+1}C_r$
$^{50}C_4+\sum\limits ^{6} _{r=1}$$^{56-r}C_3$
$=^{50}C_4+^{55}C_3+^{54}C_3+^{53}C_3+^{52}C_3+^{51}C_3+^{50}C_3$
$=(^{50}C_4+^{50}C_3)+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3$
$=(^{51}C_4+^{51}C_3)+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3$
$=(^{52}C_4+^{52}C_3)+^{53}C_3+^{54}C_3+^{55}C_3$
$=(^{53}C_4+^{53}C_3)+^{54}C_3+^{55}C_3$
$=(^{54}C_4+^{54}C_3)+^{55}C_3$
$=^{55}C_4+^{55}C_3=^{56}C_4$

 

answered Aug 13, 2013 by rvidyagovindarajan_1
edited Aug 13, 2013 by rvidyagovindarajan_1
 

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