How many odd five-digit counting numbers can be formed by choosing digits from the set $\{1, 2, 3, 4, 5, 6, 7\}$ if digits can be repeated?
7x7x7x7x4 as tertre said because only the last digits matters, as 1, 3, 5, and 7 are the odds of the group.
The place values can be anything, anything, odd
7*7*4=196 odd-five digit numbers.
1 - 7^4 x 4 =9604 Seven digits that can be repeated and which end in one of 4 odd digits [1, 3, 5, 7]
2 - [7^5 x 4] /7 =9604 Five out of 7 digits that be can be repeated that can only end in [1, 3, 5, 7] out of 7.
+-273 
+4601 
