1. ## Factorials

what is the largest two-digit prime factor of $\displaystyle 200!$ divided by $\displaystyle (100! 100!)$

2. you have to cancel this one down, not sure if I can get the answer but I can kick it off for you....

$\displaystyle \frac{200!}{100!\times 100!}$

$\displaystyle \frac{200\times 199\times 198 \times \cdots101 \times 100!}{100!\times 100!}$

$\displaystyle 100!$ will cancel

$\displaystyle \frac{200\times 199\times 198 \times \cdots 101 }{100!}$

expanding the bottom

$\displaystyle \frac{200\times 199\times 198 \times \cdots 101 }{100\times 99\times 98 \times \cdots 1 }$

Now first 1st term of 2 should cancel on the bottom

$\displaystyle \frac{2\times 199\times 2 \times \cdots 101 }{ 99\times 97 \times \cdots 1 }$

There is 50 of them

$\displaystyle \frac{2^{50} \times 199\times 197 \times \cdots 101 }{ 99\times 97 \times \cdots 1 }$

leaving the bottom in the form

$\displaystyle \frac{2^{50} \times 199\times 197 \times \cdots 101 }{ 99!! }$

not sure where to go from here...