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

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

0 0 0
will this method work for C(n, r) ?
this works for any C (n,r). why not.
will you explain with the steps for C(12, 5), C(1024, 64) and C(100, 37)
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.
0 0 0
If C(1024, 64) is to be determined as even or odd then will this approach work ??
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