Question

EVEN-ODD

Let $P(n, r) = \frac{n !}{(n - r)!}$

Given n = $2^{10}$ and r = $2^{6}$

Determine whether P(n, r) is Even or Odd.

You must provide the steps involved in solving the problem

Answer 1

  n !     
--------          =     n (n-1) (n-2) ...  (n-r+1)
  (n-r)!               

1024 * 1023 * 1022 .... 960
Clearly it is Even.
Another way: Product of two consecutive Integers is always EVEN.

Answer 2

$P(n;\ r)=\frac{n!}{(n-r)!}\\\\n=2^{10};\ r=2^6\\\\P(2^{10};\ 2^6)=\frac{2^{10}!}{(2^{10}-2^6)!}=\frac{1024!}{(1024-64)!}=\frac{1024!}{960!}=\frac{960!\cdot961\cdot962\cdot...\cdot2024}{960!}\\\\=961\cdot962\cdot...\cdot2024\\\\The\ product\ in\ which\ there\ is\ an\ even\ number\ is\ even\\\\Conclusion:P(2^{10};\ 2^6)\ is\ even.$