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:

A palindrome is a number that is the same when reading in both ways (right to left, and left to right), for example, 121

Then, we have 9 digits, and all the digits need to be even.

the options are: 0, 2, 4, 6, 8.

Now, we can tink a 9 digit number as 9 empty slots, and in each slot, we can put a number.

But because this is a palindrome, the first digit must be equal to the ninth, and the second digit must be equal to the eight, and so on.

So we can tink it as actually only 5 slots, where in each slot, we can put an even number, now let's count the options that we have in each selection.

For the first digit we have 4 options: 2, 4, 6 and 8 (0 is not counted here because if the first digit was a 0, then this would not be a 9 digit number).

for the second digit, we have also 4 options (because we already toked one, but now the 0 can be chosen)

for the third digit, we have 3 options

for the fourth digit, we have 2 options

for the fifth digit, we have only one option.

The total number of combinations is equal to the product of the number of options for each selection:

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

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58