How many 4-digit numbers can you build if the number must be odd odd digits can be repeated and even digits cannot be repeated?

Answer 1

Given (reworded a little):

• Length (L) = 4
• Must be an odd number is the same as saying it must end in an odd number
• Odd digits: 1, 3, 5, 7, 9 (nodd = 5)
• Even digits: 0, 2, 4, 6, 8 (neven = 5)
• n = Odd + Even = 10
• Odd digits can be repetitive
• Even digits cannot be repetitive
• Assuming we are not counting numbers that start with a 0, because they simplify into three digit numbers. (i.e., 0123 = 123)

First figure how how many four digit numbers in total you can have. This optional step gives you a check for your final answer, to make sure it makes sense.

I'll use square brackets to illustrate what can be in the boxes.

[n-1][10][10][10] = (9)(103) = 9000 total possibilities (this makes since because 9999 = 9000)

Because our number has to be odd:

[n-1][10][10][nodd] = (9)(5)(102) = 4500 total odd possibilities

Any number over 4500 does not make since for a final answer.

Break this into cases to simplify it.

Case 1: Number starts with an even number (remember if it starts with an odd number, we need to exclude zero). The first term will always be neven-1, but the next term, 0 is allowed, but you still have to remove the number you just used so it will also be neven-1.

EEEO => [neven-1][neven-1][neven-2][nodd] = 4 x 4 x 3 x 5 = 240

EOEO =>[neven-1][nodd][neven-1][nodd] = 4 x 5 x 4 x 5 = 400

EEOO =>[neven-1][neven-1][nodd][nodd] = 4 x 4 x 5 x 5 = 400

EOOO =>[neven-1][nodd][nodd][nodd] = 4 x 5 x 5 x 5 = 500

Case 1 total = 240 + 400 + 400 + 500 = 1540 possibilities

Case 2: Number starts with an odd number

OOOO => [nodd][nodd][nodd][nodd] = 5 x 5 x 5 x 5 = 625

OEOO =>[nodd][neven][nodd][nodd] = 5 x 5 x 5 x 5= 625

OOEO =>[nodd][nodd][neven][nodd] = 5 x 5 x 5 x 5= 625

OEEO =>[nodd][neven][neven-1][nodd] = 5 x 5 x 4 x 5 = 500

Case 2 total = 625 + 625 + 625 + 500 = 2375 possibilities

Add Case 1 and Case 2 totals up:

1540 + 2375 = 3915 possibilities

Check your answer against all odd number possibilities:

3915 < 4500, therefore our answer makes sense.

The answer is 3915 possibilities.

To check your work, try to find how many odd numbers have even numbers that repeat and subtract that from the total odd possibilities we found first. In this case though, this check is harder than the actual problem. You have to consider the EEEO, EOEO, EEOO, and OEEO cases. However the EEEO case breaks into three further cases, depending on which E is repeating.