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In a particular gumball machine, there are 4 identical blue gumballs,

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In a particular gumball machine, there are 4 identical blue gumballs, [#permalink]

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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?

A. 105
B. 210
C. 315
D. 420
E. 630
[Reveal] Spoiler: OA

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Re: In a particular gumball machine, there are 4 identical blue gumballs, [#permalink]

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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 = \(\frac{7!}{4!*2!*1!} = \frac{7*6*5}{2} = 105\)

So total arrangements of the 10 gumballs, with all 3 reds chosen last is 105 *1 = 105

Answer: A
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Re: In a particular gumball machine, there are 4 identical blue gumballs, [#permalink]

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New post 09 Aug 2017, 13:24
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Re: In a particular gumball machine, there are 4 identical blue gumballs,   [#permalink] 09 Aug 2017, 13:24
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