canbtg wrote: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 colors. How many different arrangements are possible?
A)30
B)60
c)120
D)240
E)480
OA
B
Red Marbles:
There are only TWO ways to keep the 5 red marbles separated from each other such that the marbles on the ends are of different colors:
Case 1: R _ R _ R _ R _ R _
Case 2: _ R _ R _ R _ R _ R
Number of options = 2.
Other marbles:
The two cases above guarantee that neither the 2 blue marbles nor the 2 green marbles will be in adjacent slots.
Thus, the 5 remaining positions in each case can accommodate ANY arrangement of the 5 remaining marbles.
The number of ways to arrange 5 distinct elements = 5!.
But the 5 remaining marbles are NOT distinct: they include 2 identical blue marbles and 2 identical green marbles.
When an arrangement includes IDENTICAL elements, we must divide by the number ways each set of identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the 2 identical blue marbles and by another 2! to account for the 2 identical green marbles:
Number of options = 5!/(2!2!) = 30.
To combine the options above, we multiply:
2*30 = 60.
The correct answer is
B.
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