Question

The sum ${{}^{26}C}_6+{\sum }_{r=0}^5{{}^{21+r}C}_5$ is equal ${{}^{n}C}_6$. The value is

• A. $570976$
• B. $471976$
  
• C. $681976$
  
• D. $475176$
  

Answer 1

The correct option is B $471976$

Given that

${}^{26}{C}_6+{\sum }_{r=0}^5{}^{21+r}{C}_5{=}^n{C}_6$

Now expand this

${}^{26}{C}_6{+}^{21}{C}_5{+}^{22}{C}_5{+}^{23}{C}_5{+}^{24}{C}_5{+}^{25}{C}_5{+}^{26}{C}_5{=}^n{C}_6$

$\frac{26!}{6!(26-6)!}+\frac{21!}{5!(21-5)!}+\frac{22!}{5!(22-5)!}+\frac{23!}{5!(23-5)!}+\frac{24!}{5!(24-5)!}+\frac{25!}{5!(25-5)!}+\frac{26!}{5!(26-5)!}=\frac{n!}{6!(n-6)!}$

$\frac{26!}{6!\left(20\right)!}+\frac{21!}{5!\left(16\right)!}+\frac{22!}{5!\left(17\right)!}+\frac{23!}{5!\left(18\right)!}+\frac{24!}{5!\left(19\right)!}+\frac{25!}{5!\left(20\right)!}+\frac{26!}{5!\left(21\right)!}=\frac{n!}{6!(n-6)!}$

$\frac{26!}{6!\left(20\right)!}+\frac{21!}{5!\left(16\right)!}+\frac{22!}{5!\left(17\right)!}+\frac{23!}{5!\left(18\right)!}+\frac{24!}{5!\left(19\right)!}+\frac{25!}{5!\left(20\right)!}+\frac{26!}{5!\left(21\right)!}=\frac{n!}{6!(n-6)!}$

$230230+20349+26334+33649+42504+53130+65780=\frac{n!}{6!(n-6)!}$

$\frac{n!}{6!(n-6)!}=471976$