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

b

2

−

b

b

3

+

7

b

2

+

14

b

+

6

%

−

3

b

3

+

7

b

2

+

14

b

+

6

%

step-by-step explanation:

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )