A bag contains ten marbles of the same size: 3 are identical green mar

Question

A bag contains ten marbles of the same size: 3 are identical green marbles, 2 are identical red marbles, and the other 5 are five distinct colors. If 5 marbles are selected at random, how many distinct combinations of five marbles could be drawn?
(A) 41
(B) 51
(C) 62
(D) 72
(E) 82

Counting & combination problems on the GMAT are tricky because many of them don't fit into nice neat formulas. They require some efficient out-of-the-box thinking, as this problem does. For a set of challenging counting problems, as well as for the OE of this particular problem, see:
https://magoosh.com/gmat/2015/counting- ... -the-gmat/

Mike :-)

chetan2u · Accepted Answer

hi..
 There will be different cases, which we will take one by one.

1)  Case 1 : When all five colours are different
 7C5=\frac{7*6}{2}=21

2)  Case 2 : When there are 4 different colours out of which one colour has two marbles. 
Choose 3 marbles of different colours in 6C3 ways and 2 of same colour can be only from green or red, that is 2C1
 6C3*2C1=\frac{6*5*4}{3*2*1}*2=40

3)  Case 3 : When there are 3 different colours out of which one colour has three marbles. 
Choose 2 marbles of different colours in 6C2 ways and 3 of same colour can be only from green, that is 1C1
 6C2*1C1=\frac{6*5}{2*1}*1=15

4)  Case 4 : When there are 3 different colours out of which two colours have two marbles each. 
Choose 1 marble from 5 different colours that have just one marble each in 5C1 ways and 2 each of same colour can be only from green and red, that is 2C2.
 5C1*2C2=\frac{5}{1}*1=5

5)  Case 5 : When there are two different colours and only combination is 3 of green and 2 of red. 
1 way

Total:  21+40+15+5+1=82 ways

E

thefibonacci · Answer

3G , 2R , and 5 Distinct.

All different --> C(7,5) = 21 
2 same and 3 diff --> C(2,1)*C(6,3) = 40 
3 same and 2 diff --> C(6,2) = 15 
1 from distinct set + 2 green + 2 red --> C(5,1) = 5
3 green & 2 red --> 1

total = 82.

good question! :D

akhil911 · Answer

Hi Mike,

I have seen the solution on the Magoosh website but find that it is a very cumbersome solution.
Is there any shorter way to solve this problem ???

mikemcgarry · Answer

Dear akhil911 My friend, I'm not sure you realize the inherent problem with the question you were asking. You are an intelligent individual, so I believe if you fully appreciated what the question implied, you wouldn't have asked it this way. I and all the other GMAT experts from other prep companies --- we are all educators. We are concerned with teaching, with communicating material as efficiently as possible. If we write solutions to a problem, we are concerned with showing clearest and most efficient solutions. We would have absolutely no motivation to show a complicated solution if a much simply one were possible. Therefore, do you see the only possible implication if the solution one of us shows is, in fact, very complicated? Since we are educators striving to make everything as clear and as brief as possible, and since the solution shown, for example, in this problem, is not simple, don't you see how this inescapably has to reflect something about the nature of the mathematics itself? That problem is, in some ways, representative of a 700-level Quant question. There is no simple formula one can use, just plug two or three things in a done. One has to map out a plan of attach to move through the problem. In fact, in the solution, I demonstrated a systematic way to move through the cases: if you were adept in your analysis, you could move through that entire solution in under 1 min. Producing and executing that kind of analysis: that's precisely the kind of creativity and mental flexibility you need to have if you want to cross the 700 threshold. Does all this make sense? Mike :-)

akhil911 · Answer

Hi Mike,

I Apologize if you found my post was not appropriate in this regard.
I am one of your biggest fans and appreciate the way you teach all the fellow students.
I had understood what the problem was asking and that you were trying to explain the inherent concept underlying the question.
I found the post really useful for me.
The only question I had regarding this question that is there a shorter way to approach such problems as it would take up quite a lot of time if we started counting in such questions on the real Gmat.
I could not think of one in this case and that's the reason I approached you.
Kindly accept my Apologies.

Regards
Akhil

mikemcgarry · Answer

Dear Akhil, Thank you for your kind words. I took no offense. I guess I would say --- understand that the GMAT will never expect you to do some kind of laborious counting that would take 5-10 minutes. BUT, the GMAT may expect you to dissect a problem quickly, and realize that a sum can be assembled relatively quickly from simple separate cases. I realize if this way of thinking is new to you, it may seem like a cumbersome and time-consuming approach, but what I am suggesting is that, with practice, one can master this kind of thinking and get fast at it. Does this make sense? Mike :-)

PerfectScores · Answer

So we have GGGRRABCDE and we have to select five such that all the combinations are distinct:

Let us say X denotes distinct then we have the following combinations:
GGGXX, GGXXX, GXXXX, XXXXX

GGGXX = GGGXX + GGGRR
GGXXX = GGXXX + GGRRX
GXXXX = GXXXX + GRRXX
XXXXX = RRXXX + XXXXX

GGGXX = 1 x (6C2) = 15
GGGRR = 1 way
GGXXX = 1 x 6C3 = 20
GGRRX = 5
GXXXX = 6C4 = 15
GRRXX = 5C2 = 10
RRXXX = 5C3 = 10
XXXXX = 6C5 = 6

Total number of ways = 82

tgubbay1 · Answer

Hey Mike,

Thanks for the question! I think I'm pretty good at the quant section, but combinatorics really stifle me! I can't seem to identify the technique that I have to use. I understand the explanations completely, but they're things I can never come up with on my own.

What do you recommend in this case? I'm very intimidated by Counting problems; Is there a way around this? Is there a source that outlines different methods to be applied in different cases? Please help!

Thanks a lot!

mikemcgarry · Answer

Dear tgubbay1 , I'm happy to respond. :-)

Counting problems, combinatorics, are not really about techniques. Much more important is the ability to frame the problem correctly, and this is a right-brain skill, a kind of pattern-matching that can't be summarized in a step-by-step way. See the discussion in this blog article: https://magoosh.com/gmat/2013/difficult- ... -problems/ Mike :-)

sriramkrishnan · Answer

DDDDD = 1
RDDDD -> 5C4 = 5
GDDDD -> 5C4 = 5
RRDDD -> 5C3 = 10
GGDDD -> 5C3 = 10
GGGDD -> 5C2 = 10
RGDDD -> 5C3 = 10
RGGDD -> 5C2 = 10
RGGGD -> 5C1 = 5
RRGDD -> 5C2 = 10
RRGGD -> 5C1 = 5
RRGGG = 1

Combinations are combinations of D. Hence total is 82 combinations.

PMZ21 · Answer

Hi Mike,

Please correct me if I'm wrong, as I'm slightly confused with what the question stem is asking.

When the question states - " If 5 marbles are selected at random, how many distinct combinations of five marbles could be drawn?" I inferred it as how many such combinations can we get, where each of the marbles chosen are distinct (by color, ofc).

And so my solution was: 5C5 + 5C3 (w/ one green and one red) + 5C4*2 (4 distinct with one green + 4 distinct with one red) = 21
Could you please tell me where have I gone wrong? I have gone through your solution in the blog post. However, that is taking into consideration 2 greens etc - which would mean that the 5 chosen marbles are not distinct. Please help!

mikemcgarry · Answer

Dear PMZ21 I'm happy to respond. :-)

My friend, what I wrote was: . . . how many distinct combinations of five marbles . . . and what you interpreted was: . . . how many combinations of five distinct marbles . . . If you compare those two, you will see that in my question, the combinations , not the marbles are distinct. Let's say the colors are green = G red = R black = B white = W yellow = Y purple = P sky blue = S What I am saying is that {G, G, R, R, W} is a distinct combination from {G, G, R, R, P}. In other words, if we list the colors of the five marbles, that overall list of colors will be different for different combinations. It's not that each marble will be a different color. BTW, for what you were calculating, how many different sets of five in which each of the five marbles is a different color, I think you calculated it the hard way. You could just use 7C5 = 21. Does all this make sense? Mike :-)

PMZ21 · Answer

Hi Mike, Perfect, yes that makes complete sense. Thanks! You and Bunuel are such a blessing

Posted from my mobile device

EDAXavier · Answer

Sorry, would you explain a little bit more about how you calculate case 2 and case 4? 
because the result I got for case 2 is 40 and for case 4 is 5. 
Thank you

bandook · Answer

My process was long but I hope it is a bit more easier to understand. First I found the various distinct combinations as such -
- - Note abbreviations --
-> 3 Green Marbles = 3G
-> 2 Red Marbles = 2G
-> 5 Distinct Marbles = 5D

Combinations -
1) No of ways of selecting 5, when 5 selected from 5D =  1  way

2) Lets consider situations when 4 are selected from the distinct colored
(4D+1G) and (4D+1R) = 5C4 x 2 =  10  ways

3) No of ways of selecting 5, when 3 selected from D group
(3D+2G) and (3D+2R) and (3D+1G+1R) = 5C3 x 3 =  30  ways

4) No of ways of selecting 5, when 2 selected from D group
(2D+3G) and (2D+2G+1R) and (2D+1G+2R) = 5C2 x 3 =  30  ways

5) No of ways of selecting 5, when only 1 selected from D group
(1D+3G+R) and (1D+2G+2R) = 5C1 x 2 =  10  ways

6) No of ways of selecting 5, when 0 selected from D group
Only one way - (3G+2R) =  1  way

Total = 82 ways

Doing it this method is not really good as you can miss some cases, but it helps understand whats happening. For example, lets consider the 5 distinct colours are Violet, Indigo, Blue, Yellow, Orange and White i.e. {V,I,B,Y,O,W}

So in point 2 described above - each 4D combination can have the last marble as G or R, simply put
a single combination of 4D (such as V,I,B,Y) has 2 ways to be picked either with Green or Red = {V,I,B,Y,R} or {V,I,B,Y,G}. Hence in this case, each 4D combination is multiplied by 2.

Similarly in point 3, for a 3D combination, the last 2 marbles can be picked in 3 ways, i.e for a single 3D combination the following combinations exist -
{V,I,B,G,G}
{V,I,B,G,R}
{V,I,B,R,R} = 3 ways for {V,I,B} to be completed. Hence we multiply each 3D combination by 3.

and so on...

Crytiocanalyst · Answer

Chosing all different colours :-7c5=21

Chosing 3 different colour 2 of the same colour =6c3*2=30

Chosing 2 different colours and 3 of the same colour =6c2=15

Chosing 1 different colours and 2 of two different colors=5c1*4!/2!2!=15

Chosing of one type and 2 of other colour=1
Total = 82 IMO E

Fdambro294 · Answer

Really hard question (or unique at the least)

3 identical green: it does not matter which green we pick, they are all the same. So any 2 of the Green picked will be the same combination as any other 2 of the Green picked 
G- G - G

2 identical red: same logic 
R - R - R

5 different colors: call them —- A - B - C - D - E

(1st) need to set up the different cases in which we can have multiple of the same color, without over counting and making sure to cover every distinct possibility

Case 1: 3 G——- 2 R

1 way

Case 2: 3 G ——— the other 2 are unique

(1 way for 3 green) 
and
R, A, B, C, D, E, —- from which we need to pick 2 more

(6 c 2)

Case 3: 2 G —— 2R —— and 1 more picked from 5 different colors

(5 c 1)

Case 4: 2 G —- and 3 picked from 6 of the other colors: R, A, B, C, D, E

(6 c 3)

Case 5: 2 R —— and 3 picked from 6 of the other colors: G, A, B, C, D, E

(6 c 3)

Case 6: every one of the 5 colors picked is unique:
G , R , A , B , C , D , E

(7 c 5)

(2nd) Sum up all the number of unique color combinations from each case

1 + (6 c 2) + (5 c 1) + (6 c 3) + (6 c 3) + (7 c 5) =

1 + 15 + 5 + 20 + 20 + 21 =

82 ways
E

Posted from my mobile device

Mugdho · Answer

chetan2u  
in step 2, how did we get 6C3*2C1?
shouldn’t it be 5C3 to select 3 different from 5 colors ? can u elaborate it little more plz?

Posted from my mobile device

KarishmaB  Help Please

chetan2u · Answer

There are 5 distinct colors, R and G, so total 7 different colors. 
We are looking for cases where we select 2 of same colour and rest 3 are are distinct.

Now, 2 of same colour means it can be either R or G, so 2C1 ways. 
What are we left with now: one of R or G and 5 distinct colours, that is 6 different/distinct colours. Choose 3 out of them in 6C3 ways.

A bag contains ten marbles of the same size: 3 are identical green mar

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