Combination problem with 4 groups of values

[chiraganand]

Summary:: 10 values are divided into 4 groups and need a combination of these with restrictions placed on group size, ordering and combinations

I have a combinations question.. i have 4 group of values A , B, C and D, with A-2 values, B-3 values , C-2 values, D-3 values.
1. From each group only one value has to be taken A and B are together and C and D are together. So 1 value from A , 1 Value from B and then different combinations of C and D (maintaining 1 value each).
2. The order of the values from the groups is interchangeable that is 1-A 1-B 1-C 1-D or 1-B 1-A 1-C 1-D or 1-B 1-A 1-D 1-C so on..
I am trying to calculate the number of combinations but I am coming up short.. according to me there should be 36 different combinations but i am unable to take the flexible ordering into account. Can someone help me with this

[Moderator's note: moved from a technical forum.]

 

[anuttarasammyak]

The ways to group ten all different elements into A,B,C and D count
$$_{10}C_3\ \ _7C_3\ \ _ 4C_2=25,200$$
I was not able to understand what you mean to go further. I appreciate if you show examples of your operation on the set e.g.
$$\{0,1,2,3,4,5,6,7,8,9\}$$

 

[chiraganand]

Thanks for the reply, so for example A={0,1}, B={2,3,4}, C={5,6}, D={7,8,9}. So one combination would be 0257 0258 0259 0267 0268 0269, and so on with 03.., 04.. 12..,13..14.. and also 2075 or 31 76 .. and so on.. i hope i am making sense now? The elements from A and B have to take the first two places (1st and 2nd or 2nd and 1st) and the elements of C and D have to take the 3rd and 4th places(can also be 4th and 3rd)

 

[FactChecker]

The number of ways in the order ABCD: 2*3*2*3=36
The number of ways where A and B can be swapped: 2*36=72
The number of ways where C and D can also be swapped: 2*72=144
That is all I can understand from your description of the problem.

 

[anuttarasammyak]

Thanks.

so for example A={0,1}, B={2,3,4}, C={5,6}, D={7,8,9}.

The 36 combinations are
0257
0258
0259
0267
0268
0269
0357
0358
0359
0367
0368
0369
0457
0458
0459
0467
0468
0469
1257
1258
1259
1267
1268
1269
1357
1358
1359
1367
1368
1369
1457
1458
1459
1467
1468
1469
Is this helpful ?

 

[chiraganand]

The number of ways in the order ABCD: 2*3*2*3=36
The number of ways where A and B can be swapped: 2*36=72
The number of ways where C and D can also be swapped: 2*72=144
That is all I can understand from your description of the problem.

Yes this was the answer which I came up with later. I was wondering if there is a combination formula for the same

 

[chiraganand]

Thanks.

The 36 combinations are
0257
0258
0259
0267
0268
0269
0357
0358
0359
0367
0368
0369
0457
0458
0459
0467
0468
0469
1257
1258
1259
1267
1268
1269
1357
1358
1359
1367
1368
1369
1457
1458
1459
1467
1468
1469
Is this helpful ?

Yes i understand the combinations but as the number of elements increases, how do i calculate the total number of combinations

 

[anuttarasammyak]

After you choose a division you get 36 combinations as you understand.
With all the possible divisions you get $_{10}C_4$ cases.
Is it helpful ?