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

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.