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

Question

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

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

BillyZ · Accepted Answer

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  4 ∗ 3 ∗ 2 = 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.

Bunuel · Answer

henrii · Answer

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!

vituutiv · Answer

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

Thanks!

ShankSouljaBoi · Answer

Identical objects division into r no of people is given n+r-1Cr-1 or by |000 00 0 {one separator(means two people or two sub groups or two divisions) and 6 identical objects}
Lets break this into smaller chunks
1 division of 3 identical red marbles within two people. |ooo ----> This can be done in 4!/3! ways = 4 ways
2 division of 2 identical green marbles within two people. |oo ----> This can be done in 3!/2! ways = 3 ways
3 division of 1 green marble within two people. This is the easiest,it can be with either Nora or Med. So only two possibilities.

Required =1*2*3 = 4*3*2 = 24

Thelegend2631 · Answer

Why isn't this? RRRGGB| Arranging of the above temrs? Which will b 7!/3!*2! 420.. What's happening

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IanStewart · Answer

You might imagine some rearrangements you'd be producing that way, and think of what they mean in the context of this question. For example, you might get this arrangement:

RGR | RGB

which would mean "Mel gets red, green and red marbles" and Nora gets the rest. But when you count arrangements as you're doing, you'll get a different arrangement that looks like this:

GRR | BGR

which means "Mel gets green, red and red marbles" and Nora gets the rest. But that's the same situation we counted a minute ago, and we don't want to count it again.

When you arrange letters to make a word, you're automatically assuming order is important, because the only thing that makes one word different from another that uses the same letters is the order those letters are in. In this problem, we aren't handing out marbles to the people in order, so we don't want to count as if order is important. When you do, you end up inadvertently double-counting dozens of things, and you'll get an answer that is much too large.

Posted from my mobile device

Niveditha28 · Answer

Bunuel   VeritasKarishma  I took 6 marbles in total and worked with the cases (0 ,6) ,(1,5), (2, 4) , (3 ,3) .This way I was getting 74 as total no. of ways marbles can be distributed

Please correct me why this logic is incorrect. Why does the color of the marble matter when we are distributing marbles between two people?

KarishmaB · Answer

When we say that there are 3 red marbles, the 3 marbles are identical. So if Mel gets 1 red marble. it doesn't matter which one it is.

Even if you use the cases 
(0 ,6) - 2 ways (Mel or Nora gets 0)
(1,5) - 3 ways in which 1 marble can be chosen (red/green/blue) and given to either Mel or Nora. 6 ways
(2, 4) - 2 ways of choosing same color marbles and 3 ways of choosing different color marbles. Giving 2 to either Mel or Nora. Total 10 ways
(3 ,3) - 1 way of choosing all same colour, 2 ways of choosing two marbles of same colour * 2 for the second color, 1 way of choosing all three different color marbles. Total 6 ways

In all, 24 ways.

Niveditha28 · Answer

VeritasKarishma  while i understand the logic you explained above , I'm still not clear why my approach is wrong . I used the combinations route of solving. The balls can be divided in 4 different combinations as listed below:

(0,6) or (6,0) - 6C6 *( 2 ways ) = 2
(1,5) or ( 5 ,1) - 6C1 *5C5 * (2 ways) =12
(2,4) or (4 , 2) - 6C2 * 4C4 * 2 ways=30
(3,3) - only 1 way - 6C3*3C3 = 20
 all 4 combinations sum up to 64 .

KarishmaB · Answer

Niveditha28  - Think about what this means.

(0,6) or (6,0) - 6C6 *( 2 ways ) = 2
What is (0, 6) or (6, 0) - These are the 2 ways in which all 6 marbles go to either Mel or Nora.

(1,5) or ( 5 ,1) 
The number of ways in 1 marble goes to Mel or Nora. Can you do 6C1 here? What does 6C1 represent? The number of ways in which you can select 1 out of 6 distinct objects. Do we have 6 distinct objects? No. We can 3 identical, 2 identical and 1 other. So we can select 1 object in only 3 ways. When you pick a red marble, is it any different from the other 2 red marbles? No. 
Similarly, you cannot use 6C2 and 6C3.

Rod728 · Answer

Bunuel, Karishma, I solved it in a different way than previous answers, can you check my reasoning pls?

I picked Mel and analyzed the possibilities of getting 0, 1, 2 or 3 marbles.

If he gets 0 marbles = 1 way
If he gets 1 marble = 3 ways (He could get either R, G or B)
If he gets 2 marbles= 5 ways (RR, GG, RG, RB, GB)
Now I multiplied by 2 because there is a similarity for getting 4, 5 and 6 marbles. So it is =  2 (1+3+5) = 18. 
Finally, if he gets 3 marbles=  6 ways  (3R, 2R1G, 2G1R, RGB, 2R1B, 2G1B). There is no similarity here because both Mel and Nora get 3.
 18+6=24 ways >>> answer C.

Pragun1710 · Answer

Can this problem be solved by using the below formula https://gmatclub.com/forum/download/file.php?id=83071 GMAT-Club-Forum-zwz59ip2.png

Wadree · Answer

Lets think bout something......there are two cakes and two people A and B ...... so how many possibility if its not necessary for them to get at least 1 cake?...... 3 possiblity......   ....
  .....   ....
So.... If there are H number of things ... and ... its not necessary to get at least one thing ... then there are H + 1 number of scenarios to arrange them between 2 people.......
So....number of scenarios to arrange 1 blue marble among mel and sora .... 1 + 1 = 2 .....
number of scenarios to arrange 2 green marble among mel and sora ..... 2 + 1 = 3 .....
number of scenarios to arrange 3 red marble among mel and sora .... 3 + 1 = 4 ......
Total scenarios ..... 2 × 3 × 4 = 24 .......

! nah id win!

shivani1351 · Answer

We can use the popular Beggar's Method individually for all the three types of marbles here:

For the Red Marbles -> 3+2-1 C 2-1, i.e. 4C1 = 4
For the Green Marbles -> 2+2-1 C 2-1, i.e. 3C1 = 3
For the Blue Marble -> 1+2-1 C 2-1, i.e. 2C1 = 2

Since we need to distribute them all between Mel and Nora:

Ans = 4*3*2 = 24

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

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