Answer:
Consider lining up all the blue marbles and placing one red marble in between them. Let that arrangement be fixed. The remaining red marbles will be 5 red marbles. Note that the arrangement will be:
x BR x BR x BR x BR x BR x BR x BR x B x
Note that this is the same even if our arrangement is
x B x RB x RB x RB x RB x RB x RB x RB x
Note that there are 9 spots, denoted by x, where we can put the 5 remaining red marbles. To find the number of ways of putting the remaining 5 red marbles to x, it is similar in finding the number of non-negative solution for the equation
![a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 = 5](https://tex.z-dn.net/?f=a_1%20%2B%20a_2%20%2B%20a_3%20%2B%20a_4%20%2B%20a_5%20%2B%20a_6%20%2B%20a_7%20%2B%20a_8%20%2B%20a_9%20%3D%205)
which is given by
= 1287
Hence, there are 1287 ways of arranging 8 blue marbles and 12 red marbles without placing 2 blue marbles next to each other.
Answer:
(12,-6)
Step-by-step explanation:
I hope this helps and have a good day.
I believe that would be N - 8, N being the number