The value of ^50C_4 + _r=1^6^56-rC_3 is:

Question

The value of $^{50}C_4 + \displaystyle\sum_{r=1}^{6}{}^{56-r}C_3$ is:

Accepted Answer

Given, $${}^{50}{C_4} + \sum\limits_{n = 1}^6 {{}^{56 - r}{C_3}} $$

$$ \Rightarrow $$ ${}^{50}{C_4}$ + ${}^{55}{C_3}$ + ${}^{54}{C_3}$ + ${}^{53}{C_3}$ + ${}^{52}{C_3}$ + ${}^{51}{C_3}$ + ${}^{50}{C_3}$

Arrange those this way

$$ \Rightarrow $$ ${}^{50}{C_4}$ + ${}^{50}{C_3}$ + ${}^{51}{C_3}$ + ${}^{52}{C_3}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$

We know this formula [ ${{}^n{C_r}}$ + ${{}^n{C_{r - 1}}}$ = ${{}^{n + 1}{C_r}}$ ] which is used to solve this problem.

$$ \Rightarrow $$ ${}^{51}{C_4}$ + ${}^{51}{C_3}$ + ${}^{52}{C_3}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$

$$ \Rightarrow $$ ${}^{52}{C_4}$ + ${}^{52}{C_3}$ + ${}^{53}{C_3}$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$

$$ \Rightarrow $$ ${}^{53}{C_4}$+${}^{53}{C_3}$+${}^{54}{C_3}$+${}^{55}{C_3}$$ \Rightarrow$$ $ {}\^{54}{C\_4}$$ + ${}^{54}{C_3}$ + ${}^{55}{C_3}$

$$ \Rightarrow $$ ${}^{55}{C_4}$+${}^{55}{C_3}$$ \Rightarrow$$ $ {}\^{56}{C\_4}$$

Solutions · 1