How many 9-digit palindromes are there with all the digits being even and each digit appearing no more than twice?

96

Step-by-step explanation:

You can think a 9-digit palindrome as 5 slots.

In each slot, you can put a digit.

_ _ _ _ _

The first slot, is the 9th digit and the 1st digit.

The second slot is the 8th digit and the 2nd digit

etc.

This is for the nature of a palindrome number.

Then, we have 5 selections.

Now, remember that for the first four selections each digit will appear two times, so we can choose each even number only once.

Then:

For the first selection, we have 4 options (2, 4, 6 and 8. The zero can not be chosen here because the 9th digit can not be a zero if this is a nine digit number).

For the second selection we have 4 options, the 3 that remain for the previous selection and the 0, that is even.

for the third selection, we have 3 options.

for the second selection, we have 2 options

for the fifth selection we have only one option.

The total number of combinations is equal to the product of the number of options in each selection, this is:

C = 4*4*3*2*1 = 96

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step-by-step explanation:

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: