Six marbles (two red, two blue, and two green) are arranged in two row

The maximum number of arrangement with 1 red, 1 blue and 1 green in first row is (3!) =6
The maximum number of arrangement with 1 red, 1 blue and 1 green in second row is (3!) =6

The maximum number of arrangement with only each color appearing once in each row is 6*6 =36. i.e. first row*second row = 36

With the restrictive conditions applied of the same color not on each column
Looking at the options we already know that only option A could be the answer, because it is the only option that is less than 36. Hence option A = 12 is the answer.

There will be repetitions 2/3 of the time and no repetition 1/3 of the time. 1/3 of 36 = 12

All conditions possible are:

BRG
RGB

BRG
GBR

RGB
GBR

RGB
BRG

GBR
BRG

GBR
RGB

Let's fix the first row of marble as R B G
Possible ways to arrange the first row= 3! = 6

Second row can be- B G R or G R B
Second row can be arranged in 2 ways

Therefore, total possibilities= 6 * 2 = 12

Correct Answer= A

let the position be
XYZ
ABC

X can take any of three colours - 3
Y can take any of remaining 2 - 2
Z can take the last one - 1
A can take any of 2 except the one at X - 2
B can take ONLY 1

similarly C will take 1

total 3*2*1*2*1*1=12
A

The correct answer is option A.

Here is why:
Since no row has two marbles of the same color it can a permutation of R, B and G.
So, for the first row will be 3C1 ways i.e. 6.

Now lets assume one possiblity as R, B, G. The subsequent row can only two possible combinations i.e. G ,R, B or B, G,R

So 6 ways to arrange the first row * 2 ways to arrange the second row = 12

Hence A.

3 colors to be place in a row
3! ways
no of rows 2
so 3!x2= 12

A

There could be two approaches primarily- 1) Using combinatorics- there are 3 colors,two balls/color- RED=2,Blue=2 and Green=2

Now, there could be 2 options of choosing where to place the first ball i.e Row1 or Row2=2C1
There would be 3 options of choosing the first color- i.e. R,B,G=3C1
if we have chosen the 1st color, 2nd and 3rd color has to be different i.e. let's say if we chose Red, We would have only B and green to be selected=2C1(1 out of 2) and 1C1 subsequently

Hence total number of ways=2C1*3C1*2C1*1C1=12

2) There could be 3! possibilities of arranging a Red, a green and 1 Blue ball

RBG,RGB,BGR,BRG,GRB and GBR.

Now if we select any one of the 6 given combinations for row 1, we would only be able to select 1 out of 2 remaining combinations for Row 2( i.e for RGB in Row 1, we can select BRG and GBR only=> remaining combinations would have atleast one of the three colors in same position in column e.g. if we select GRB, Blue color will be same in 1st and 2nd row- not allowed )= 6*2=12 or 2 options for RGB OR 2Option for RBG etc.....2+2-----6 times=12

Please give kudos if it makes sense

souvik101990
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Hope this helps

Let’s let the first marble in row 1 be R. We see that one possible layout of the two rows is:

R-B-G
B-G-R

Keeping row 1 the same, the only other way to create row 2 is:

R-B-G
G-R-B

Now, let’s keep R in first position in row 1; we see that we could now have R-G-B as row 1. Using a similar strategy as we did above, we see that there are two possible ways to create row 2: either B-R-G or G-B-R.

We see that when R is in first position of row 1, there are four different marble arrangements that satisfy the stated requirements. Similarly, when G is in first position in row 1, we will obtain four different marble arrangements, and when B is in first position in row 1, we will obtain an additional four different marble arrangements.

Thus, the total number of arrangements is 4 + 4 + 4 = 12.

Answer: A

Hello,

I'm apologizing in advance if this is a stupid question, however how does this problem differ from this one:

"Three couples are seated in two rows of three seats. If no row or column of seats contains both members of a couple, how many possible arrangements are there?"

Is the fact that we're dealing with indistringuishable marbles what is driving the difference in permutations? It's not clear why we're not treating them as 6 different marbles.

Thank you,
Chris

I guess no one ever answered this question, but I will now in case anyone else has it. Yes, the two problems are different, because with 6 people, all of the people are different. In this marbles problem, the two red marbles are identical. So in the marbles problem, we only have 3 choices for the first spot in the first row, 2 for the next in the row, 1 for the last, then in the second row we have just 2 choices for the first slot (it can't match the first row), and 1 for the remaining slots, and the answer is 3*2*1*2 = 12. For the couples problem, we have 6 choices for the first slot in the first row, 4 for the next (we can't repeat the same person or use their partner), and 2 for the last, then moving to the next row we only have 2 people left who can fit the first slot, and no choice about the last two slots, and the answer is 6*4*2*2 = 96.

What if I do 6c1*4c1*2c1 for the first row? Where am I going wrong Bunuel souvik101990 KarishmaB

6C1 doesn't make sense because we don't have 6 distinct marbles. Both red marbles are the same, both blue are the same and both green are the same.
Hence you have pairs of 3 distinct marbles. Hence you don't need to make any selection. You need to put a Red, a green and a blue in the first row.
So you can arrange Red, Blue, Green in the first row in 3! ways. (Say you put them as RGB)

For the second row, below Red, you can put either Blue or Green so 2 options. Now there is only 1 way of placing the remaining two marbles.

Hence, total 3! * 2 = 12 arrangements

Answer (A)

Thanks a lot KarishmaB. This is very helpful. What if I do 2c1*2c1*2c1 for the first row. Will it be wrong? We can't do this because again the 2 marbles are not distinct? I am getting really confused with some of the questions.

Think about why you would use 2C1. Do you have to select a red marble out of given two distinct red marbles? No. Both marbles are the same. You may pick any and nothing would change. Hence no selection is required.

If you are not sure when to use permutations and when to use combinations, check my following two videos on permutations and combinations here:
https://youtu.be/LFnLKx06EMU
https://youtu.be/tUPJhcUxllQ

Given: Six marbles (two red, two blue, and two green) are arranged in two rows of three such that no single row or column has two marbles of the same color.

Asked: How many different arrangements are possible?

Let us assume the arrangement in the form: -

A B C
D E F

Number of options for A = 3
Number of options for B = 2
Number of options for C = 1
Number of options for A = 2
Number of options for A = 1
Number of options for A = 1

Total number of possible arrangements = 3*2*1*2 = 12

IMO A