Respuesta :
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
[tex]a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 = 5[/tex]
which is given by
[tex](^{5+9-1}_{9-1})[/tex] = 1287
Hence, there are 1287 ways of arranging 8 blue marbles and 12 red marbles without placing 2 blue marbles next to each other.
