In a particular gumball machine, there are 4 identical blue gumballs, 3 identical red gumballs, 2 identical green gumballs, and 1 yellow gumball. In how many different ways can the gumballs be dispensed, 1 at a time, if the 3 red gumballs are dispensed last?
Select: A) 105 B) 210 C) 315 D) 420 E) 630
Since the red gumballs are being dispensed last they will not impact the number of permutations of the other gumballs, so we can effectively ignore them. (There is only one way for the three red gumballs to be dispensed last (RRR), so we will multiply the number of permutations of the other colours by 1).   The remaining gumballs are: 4 blue, 2 green, 1 yellow. 7 gumballs in total, so they can be arranged in 7! ways. But since the gumballs of each colour are identical, we must divide by the arrangements of each colour.   Total arrangements of the 7 gumballs = 7!/4!∗2!∗1!=7∗6∗5/2=105   So total arrangements of the 10 gumballs, with all 3 reds chosen last is 105 *1= 
<gadget id="calculator">105 *1</gadget>
<output>105</output>
105   105nswer: 105
<result>A</result>
In a particular gumball machine, there are 4 identical blue gumballs, 3 identical red gumballs, 2 identical green gumballs, and 1 yellow gumball. In how many different ways can the gumballs be dispensed, 1 at a time, if the 3 red gumballs are dispensed last?