Question

The value of ${}^{47}{C}_4+\sum _{r=1}^5{}^{52-r}{C}_3$=

• A. ${}^{52}{C}_2$
• B. ${}^{52}{C}_3$
• C. ${}^{52}{C}_4$
• D. ${}^{52}{C}_5$

Answer 1

Answer: (C)

${}^{52}{C}_4$

We know that,
${}^n{C}_r+{}^n{C}_{r-1}$

$=\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}$

$=\frac{n!(n-r+1)}{r!(n-r)!(n-r+1)}+\frac{n!r}{(r-1)!(n-r+1)!r}$

$=\frac{n!(n-r+1)}{r!(n-r+1)!}+\frac{n!r}{(n-r+1)!r!}$

$=\frac{n!}{r!(n-r+1)!}(n-r+1+r)$

$=\frac{n!}{r!(n-r+1)!}(n+1)$

$=\frac{(n+1)!}{r!(n-r+1)!}$

$={}^{n+1}{C}_r$

$\Rightarrow {}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$
The equation is: $({}^{47}{C}_4+{}^{47}{C}_3)+{}^{48}{C}_3+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}{C}_3$

$=({}^{48}{C}_4+{}^{48}{C}_3)+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}{C}_3$

$=({}^{49}{C}_4+{}^{49}{C}_3)+{}^{50}{C}_3+{}^{51}{C}_3$

$=({}^{50}{C}_4+{}^{50}{C}_3)+{}^{51}{C}_3$

$=({}^{51}{C}_4+{}^{51}{C}_3)$

$={}^{52}{C}_4$

Hence, option 'C' is correct.