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Q.
$\displaystyle\sum_{r=0}^{n-p}(-1)^{p+r} \cdot{ }^n C_r \cdot{ }^{n-r} C_p \cdot 5^{n-p-r} \cdot 6^r$ is equal to
Binomial Theorem
Solution:
${ }^n C_r \cdot{ }^{n-r} C_p=\frac{n !}{r !(n-r) !} \times \frac{(n-r) !}{p !(n-r-p) !} \times \frac{(n-p) !}{(n-p) !}=\frac{n !}{p !(n-p) !} \cdot \frac{(n-p) !}{(n-p-r) ! r !}={ }^n C_p \cdot{ }^{n-p} C_r$
$\Rightarrow \displaystyle \sum_{ r =0}^{ n - p }(-1)^{ p + r } \cdot{ }^{ n } C _{ r } \cdot{ }^{ n - r } C _{ p } \cdot 5^{ n - p - r } \cdot 6^{ r } $
$\Rightarrow (-1)^{ p \cdot{ }^{ n } C _{ p }} \cdot \displaystyle\sum_{ r =0}^{ n - p }{ }^{ n - p } C _{ p } \cdot 5^{ n - p - r } \cdot(-6)^{ r }$
$\Rightarrow (-1)^{ p } \cdot{ }^{ n } C _{ p } \cdot(5-6)^{ n - p }={ }^{ n } C _{ p }(-1)^{ p + n - p } $
$\Rightarrow { }^{ n } C _{ p }(-1)^{ n }$