Question 1204522: How many odd numbers greater than 70,000 can be formed using the digits 0,1,4,7,8,9.
(a) without repetition.
(b) if repetitions are allowed.
Answer by greenestamps(13214)   (Show Source):
You can put this solution on YOUR website!
(a) without repetition....
(1) 5-digit numbers:
7xxx1
7xxx9
8xxx1
8xxx7
8xxx9
9xxx1
9xxx7
For each of those 7 cases, the number of different numbers is 4*3*2 = 24.
That's 7*24 = 168 5-digit numbers
(2) 6-digit numbers:
1xxxx7
1xxxx9
4xxxx1
4xxxx7
4xxxx9
7xxxx1
7xxxx9
8xxxx1
8xxxx7
8xxxx9
9xxxx1
9xxxx7
For each of those 12 case the number of different numbers is 4*3*2*1 = 24.
That's 288 6-digit numbers.
ANSWER (without repetition): 168+288 = 456
(b) with repetition....
The patterns for the case with repetition allowed are the same; but now for each case the number of different numbers is 6*6*6*6 = 1296.
That makes the total number of different 5- or 6-digit numbers
(7+12)(1296) = 24624
However, for the case where repetition is allowed, there are in fact an infinite number of odd numbers greater than 70000, because there is no stated restriction on the number of digits.
|
|
|
