Question

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?

PLEASE HELP.

Answer 1

 How many possibilities for 5th digit? 4 [1, 3, 5, or 7.] 
How many possibilities for 1st digit? 6 [The digit used in 5th place is gone and 0 is excluded.] 
How many possibilities for 2nd digit? 6 [Even with 0 included, 2 of the 8 are gone.] 
How many possibilities for 3rd digit? 5 [3 of 8 gone.] 
How many possibilities for 4th digit? 4 [4 of 8 gone.]  
In order of position,6x6x5x4x4 = 2880.
Hope this helps!! If not, I am so sorry:(
  

Answer 2

The number of odd five-digit counting numbers is 9604

How to determine the number of odd numbers?

The set of numbers is given as:

Set = {1, 2, 3, 4, 5, 6, 7}

Since the digits can be repeated, then the first 4 digits can be any of the 7 digits.

While the last digit can be any of the 4 odd digits i.e. 1, 3, 5 and 7

So, the number of digits is;

Count = 7 * 7 * 7 * 7 * 4

Evaluate

Count = 9604

Hence, the number of odd five-digit counting numbers is 9604

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