PLEASE HELP! a. How many 7-digit numbers have odd numbers in the odd positions (i. e. odd numbers in the first, third, fifth and seventh positions)? (2)

b. How many of these have even numbers in the even positions?

a) 625,000

b) 78,125 if 0 is counted as even or 40,000 if it is not.

Step-by-step explanation:

Represent the 7 digit number by abcdefg

Position          Allowable digits          Number of allowable digits

g                      1,3,5,7,9                          5

f                       0,1,2,3,4,5,6,7,8,9          10

e                      1,3,5,7,9                          5

d                      0,1,2,3,4,5,6,7,8,9          10

c                      1,3,5,7,9                          5

b                       0,1,2,3,4,5,6,7,8,9         10

a                      1,3,5,7,9                          5

This implies that the number of 7 digit numbers with odd digits in the odd positions is

5x10x5x10x5x10x5=625000  

For part b, count 0 as an even digit.  Then rule out 1,3,5,7,9 in positions b, d, and f. This leaves 5 possible digits in all 7 locations. So the answer is

5^7=78,125    

If 0 is not counted as an even digit, then the answer is

4^3+5^4=40000

50*10 is greater

step-by-step explanation:

if you add the zero from 10 to 50, it will give you a bigger number.