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

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation:

in the math equation 8 x 7 = 56, the 8 and the 7 are both factors of 56, since in this lesson we will use square tiles to figure out the factors of a whole find as many different arrays as you can, but follow one important rule: all the rows must have the i can make two rows of nine, so 2 and 9 are both factors of 18.

step-by-step explanation: