30 July 2011

Day 1 problem

Saturday, July 30, 2011  , ,  No comments
How many zero are the end of P(10000,13)


Solution
The formula of permutation is P(n, k)=\frac{n!}{(n-k)!}. Then P(10000, 13)=\frac{10000!}{(9987)!}
so we'll find  zero the end of 10000! and 9987!.

there are n_{1}=\left [\frac{10000}{5} \right ]=2000 integers less than 10000 which are multiples of 5,
and n_{2}=\left [\frac{10000}{5^2} \right ]=400 which are multiples of 5^2,
and n_{3}=\left [\frac{10000}{5^3} \right ]=80 which are multiples of 5^3,
and n_{4}=\left [\frac{10000}{5^4} \right ]=16 which are multiples of 5^4,
and n_{5}=\left [\frac{10000}{5^5} \right ]=3 which are multiples of 5^5,

so we get 2000+400+80+16+3=2499 zero are the end of 10000!

By the same way we get 2493 zero are the end of 9987!

So there are 2499-2493=6 are the end of P(10000,13).



Please correct me! by leave your comment . Thanks

0 comments:

Post a Comment