Question

The value of the expression $^{47}C_4+ \displaystyle \sum^5_{j = 1} \,{}^{52-j}C_3$ is

• A. $^{47}C_5$
• B. $^{52}C_5$
• C. $^{52}C_4$
• D. None of these

Answer 1

Answer: (C)

Here, $^{47}C_4+ \displaystyle \sum^5_{j = 1} \,{}^{52-j}C_3$
$= \,{}^{47}C_4+ \,{}^{51}C_3 + \,{}^{50}C_3 + \,{}^{49}C_3+ \,{}^{48}C_3 + \,{}^{47}C_3$
$= (\,{}^{47}C_4+ \,{}^{47}C_3 )+ \,{}^{48}C_3 + \,{}^{49}C_3+ \,{}^{50}C_3 + \,{}^{51}C_3$
[using $\,{}^nC_r + \,{}^nC_{r-1} = \,{}^{n+1}C_r]$
$= (\,{}^{48}C_4+ \,{a}^{48}C_3 )+ \,{}^{49}C_3 + \,{}^{50}C_3 + \,{}^{51}C_3$
$= (\,{}^{49}C_4+ \,{}^{49}C_3 )+ \,{}^{50}C_3 + \,{}^{51}C_3$
$= (\,{}^{50}C_4+ \,{}^{50}C_3 )+ \,{}^{51}C_3$
$= \,{}^{51}C_4+ \,{}^{51}C_3 = \,{}^{52}C_4$