If you have 2 distinct black playing cards and 2 distinct red playing cards, how many ways can you arrange the four cards given that the red cards can never be next to each other

I'm not 100% on the interoperation of this question...

are the two red cars out of a 52 card deck and you can try all the combinations of two red and black cards ????

for this answer i will assume that you have 4 coins two nickels and 2 quarters

and the question is " how many ways can you arrange the four coins given that the nickels can not be next to the quarters"

in that case I think the answer is 8

Step-by-step explanation:

1-N1Q1N2Q2

2-N1Q2N2Q1

3-N2Q1N1Q2

4-N2Q2N1Q1

5-Q1N2Q2N1

6-Q2N2Q1N1

7-Q1N1Q2N2

8-Q2N1Q1N2

1

step-by-step explanation:

it's the only multiple of both and the least

step-by-step explanation:

a cube has 6 sides

the probability of getting 5 on the first cube is 1/6 since only one side of the cube has 5 on it.

there are 3 sides of the cube with even number on it [2,4,6]

therefore the probability of getting even number is 3/6

for both events to happen together, the probability would be

1/6 x 3/6 = 12/36

which simplifies to 1/12