Question

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?

Answer 1

a) 625,000

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

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