Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colours. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480
For Detailed Answers..
if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.
Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.
Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.
Answer: B