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# Mel and Nora share a total of three red marbles, two green marbles, an

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Math Expert
Joined: 02 Sep 2009
Posts: 47110
Mel and Nora share a total of three red marbles, two green marbles, an  [#permalink]

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16 Feb 2017, 03:32
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Difficulty:

95% (hard)

Question Stats:

42% (01:32) correct 58% (02:00) wrong based on 112 sessions

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Mel and Nora share a total of three red marbles, two green marbles, and one blue marble. In how many ways can Mel and Nora divide the marbles between themselves, if it is not necessary for each of them to get at least one marble?

A. 6
B. 18
C. 24
D. 36
E. 72

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Math Expert
Joined: 02 Sep 2009
Posts: 47110
Mel and Nora share a total of three red marbles, two green marbles, an  [#permalink]

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14 Aug 2017, 12:11
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##### General Discussion
Intern
Joined: 10 Oct 2016
Posts: 1
Re: Mel and Nora share a total of three red marbles, two green marbles, an  [#permalink]

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17 Feb 2017, 02:57
1
Let's try this one.

Number of ways we can divide the Red marbles: MMM, MMN, MNN, NNN --> 4 ways
Number of ways we can divide the Green marbles: MM, MN, NN --> 3 ways
Number of ways we can divide the Blue marble: M or N --> 2 ways

Total number of ways: 4*3*2 = 24

C for me!
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Mel and Nora share a total of three red marbles, two green marbles, an  [#permalink]

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17 Feb 2017, 22:30
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2
Bunuel wrote:
Mel and Nora share a total of three red marbles, two green marbles, and one blue marble. In how many ways can Mel and Nora divide the marbles between themselves, if it is not necessary for each of them to get at least one marble?

A. $$6$$
B. $$18$$
C. $$24$$
D. $$36$$
E. $$72$$

Official solution from Veritas Prep.

Combinatorics questions, like this one, so often come down to a matter of approach. That is, we can tell the story of this problem in different ways, each of which leads to a valid solution, but some of which lead to far easier solutions than others.

For instance, we might view this problem as “Mel could get zero, one, two, three, four, five, or six marbles, in a variety of color combinations” and then calculate/count the number of possible combinations for each possible number of marbles. We might also rely on the symmetry between Mel and Nora to cut that workload almost in half (calculate for Mel getting zero through two, multiply by $$2$$ to account for Nora getting zero through two, then add the case of Mel and Nora each getting three). But even then, we’d be expending a fair amount of brute force effort.

If we tell the story a bit differently, though, we can make the math a lot better. Rather than focus on one of the people and their number of marbles in the aggregate, focus on one of the colors and the number of that type that one person gets. When it comes to red marbles, Mel could wind up with none, one, two, or three – that’s $$4$$ possibilities. As for green marbles, Mel could have none, one, or two$$3$$ possibilities. And the blue marble is a take-it-or-leave-it proposition$$2$$ possibilities. (We don’t have to calculate anything for Nora; she’ll just get whatever is left.) Since we’re distributing red, blue, and green, we multiply these results to find out that there are 432=$$24$$ total options.

The correct answer is $$C$$.

Did you notice this problem’s connection to the Unique Factors Trick? What we’re doing here is actually a perfect analogy – we’re taking some quantities of identical items from each of several categories, and it really makes no difference whether those items are various colored marbles or various prime factors. It makes sense that the formula is the same – the number of items in each category plus one (to cover the “choose zero” possibility), and then multiply the results.
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Joined: 24 May 2017
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Re: Mel and Nora share a total of three red marbles, two green marbles, an  [#permalink]

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14 Aug 2017, 12:05
Does anyone have more problems like this? (i.e.: with two possible framings, of which one has a much simpler math)

Thanks!
Re: Mel and Nora share a total of three red marbles, two green marbles, an &nbs [#permalink] 14 Aug 2017, 12:05
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