There are only 2 ways for an odd number of even digits: 1 even digit or all even digits. Case 1: 1 even digit There are $5 \cdot 5=25$ ways to choose the odd digits, 5 ways for the even digit, and 3 ways to order the even digit. So, $25 \cdot 5 \cdot 3=375$. However, there are $5 \cdot 5=25$ ways that the hundred\textquotesingle s digit is 0 and we must subtract this from 375 , leaving us with 350 ways. Case 2: all even digits There are $5 \cdot 5 \cdot 5=125$ ways to choose the even digits, and $5 \cdot 5=25$ ways where the hundred\textquotesingle s digit is 0 . So, $125-25=100$. Adding up the cases, the answer is $100+350=$ (D) 450 . 
How many three-digit positive integers have an odd number of even digits? 
Options:
(A) $$150$$ 
(B) $$250$$ 
(C) $$350$$ 
(D) $$450$$ 
(E) $$550$$ 