How many even (E) and odd (O) combinations?

Question

In a four digit number, how many possible combinations of odd and/or even numbers are there?

I'm not using the actual numbers themselves, but just odd and/or even.

I'm half asleep and not good at math but I'm seeing 16 possibilities.

Several examples are below, where "E" is an even number and "O" is an odd number:

EEEE

OOOO

EEEO

EEOO

EOOO

EOOE

Answer 1

Yes, there are 16 for mixed odds and evens, because there are two possibilities for each of the four digits—odd or even, and 16=2⁴. However, the 4-digit numbers will include those with leading zeroes such as 0000, 0001, 0234, for example. So, leading blanks would have to count as evens.

Comments

Hey, thanks for the fast reply. Appreciated, and glad I was correct.

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I'll add the combinations below with their opposites next to them to save space:

OOOO .. EEEE

OOOE .. EEEO

OOEE .. EEOO

OEEE .. EOOO

OEOE .. EOEO

OEOO .. EOEE

OOEO .. EEOE

OEEO .. EOOE

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Because this is a binary problem you can use binary arithmetic and counting. 0 can be used to represent even and 1 odd. The numbers 0-15 converted to binary match exactly what you present as E and O:

0000 0001 0010 0011 0100 0101 0110 0111

(    0      1       2      3       4        5       6      7   )

1000 1001 1010 1011 1100 1101 1110 1111

(  8       9      10     11     12    13     14     15  )

This is, of course, the original principle of computer logic, because a binary state is Yes/No, On/Off, Even/Odd, Positive/Negative, etc. That is, any two state system.

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Thanks again :)