Identical green marbles = 6
Identical red marbles = 2
Identical blue marbles = 4
Total marbles = 12
How many distinct sets of three can be selected?
Possible sets are :
- The selected 3 are different colors, 1 red, 1 green, and 1 blue -> only one way
- The selected 3 have 2 identical marbles and 1 different marble. I.e 2 red and 1 green, 2 blue and 1 red, 2 green and 1 blue
For all the listed ways, there are alternative combinations. i.e 2 red and 1 green or 2 green and 1 red, 2 blue and 1 red or 2 red and 1 blue e.t.c number of possible ways in this set = (3C2)*2
$$3C2=\frac{3!}{2!\left(3-2\right)!}=\frac{3\cdot2\cdot1}{2\cdot1\cdot1}=\frac{6}{2}=3$$
$$3\cdot2=6\ possible\ ways$$
- The selected 3 have the same color but since there are only 2 identical red marbles, it is only possible to select 3 blue or 3 green marbles. So, there are 2 possible ways to select from this set from the 3 possible sets above. Therefore, the total possible distinct sets of 3 that can be selected = 1 + 6 + 2 = 9
Answer = A
