Finding combinations with groups.
$begingroup$
Suppose I have four groups A, B, C, D filled with items belonging to groups:
- A
- A1
- A2
- A3
- B
- B1
- B2
- B3
- B4
- C
- C1
- C2
- D
- D1
- D2
- D3
I would like to create a combination with four items, for example:
- A1, B2, C1, D2
- A2, B1, C2, D1
- so on ...
Such that every item must be from a different group.
This would be wrong: A1, A2, B1, B2 because we cannot have two items from same group.
Is it mathematically possible to create such combinations given the groups and number of items and know how many would be there?
combinations
asked Jan 6 at 16:43
user963241user963241
17318
$endgroup$
add a comment |
$begingroup$
Suppose I have four groups A, B, C, D filled with items belonging to groups:
- A
- A1
- A2
- A3
- B
- B1
- B2
- B3
- B4
- C
- C1
- C2
- D
- D1
- D2
- D3
I would like to create a combination with four items, for example:
- A1, B2, C1, D2
- A2, B1, C2, D1
- so on ...
Such that every item must be from a different group.
This would be wrong: A1, A2, B1, B2 because we cannot have two items from same group.
Is it mathematically possible to create such combinations given the groups and number of items and know how many would be there?
combinations
asked Jan 6 at 16:43
user963241user963241
17318
$endgroup$
add a comment |
$begingroup$
Suppose I have four groups A, B, C, D filled with items belonging to groups:
- A
- A1
- A2
- A3
- B
- B1
- B2
- B3
- B4
- C
- C1
- C2
- D
- D1
- D2
- D3
I would like to create a combination with four items, for example:
- A1, B2, C1, D2
- A2, B1, C2, D1
- so on ...
Such that every item must be from a different group.
This would be wrong: A1, A2, B1, B2 because we cannot have two items from same group.
Is it mathematically possible to create such combinations given the groups and number of items and know how many would be there?
combinations
asked Jan 6 at 16:43
user963241user963241
17318
$endgroup$
Suppose I have four groups A, B, C, D filled with items belonging to groups:
- A
- A1
- A2
- A3
- B
- B1
- B2
- B3
- B4
- C
- C1
- C2
- D
- D1
- D2
- D3
I would like to create a combination with four items, for example:
- A1, B2, C1, D2
- A2, B1, C2, D1
- so on ...
Such that every item must be from a different group.
This would be wrong: A1, A2, B1, B2 because we cannot have two items from same group.
Is it mathematically possible to create such combinations given the groups and number of items and know how many would be there?
combinations
combinations
asked Jan 6 at 16:43
user963241user963241
17318
asked Jan 6 at 16:43
user963241user963241
17318
asked Jan 6 at 16:43
user963241user963241
17318
asked Jan 6 at 16:43
user963241user963241
17318
asked Jan 6 at 16:43
user963241user963241
17318
17318
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, you can. We can, in fact, count the number of such combinations. Since for the first group you have $3$ choices, for the second group you have $4$ choices, for the third group you have $2$ choices and the for the fourth group you have $3$ choices thus the number of such groups is
$$3cdot 4cdot 2cdot 3=72.$$
answered Jan 6 at 16:46
model_checkermodel_checker
4,18621931
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, you can. We can, in fact, count the number of such combinations. Since for the first group you have $3$ choices, for the second group you have $4$ choices, for the third group you have $2$ choices and the for the fourth group you have $3$ choices thus the number of such groups is
$$3cdot 4cdot 2cdot 3=72.$$
answered Jan 6 at 16:46
model_checkermodel_checker
4,18621931
$endgroup$
add a comment |
$begingroup$
Yes, you can. We can, in fact, count the number of such combinations. Since for the first group you have $3$ choices, for the second group you have $4$ choices, for the third group you have $2$ choices and the for the fourth group you have $3$ choices thus the number of such groups is
$$3cdot 4cdot 2cdot 3=72.$$
answered Jan 6 at 16:46
model_checkermodel_checker
4,18621931
$endgroup$
add a comment |
$begingroup$
Yes, you can. We can, in fact, count the number of such combinations. Since for the first group you have $3$ choices, for the second group you have $4$ choices, for the third group you have $2$ choices and the for the fourth group you have $3$ choices thus the number of such groups is
$$3cdot 4cdot 2cdot 3=72.$$
answered Jan 6 at 16:46
model_checkermodel_checker
4,18621931
$endgroup$
Yes, you can. We can, in fact, count the number of such combinations. Since for the first group you have $3$ choices, for the second group you have $4$ choices, for the third group you have $2$ choices and the for the fourth group you have $3$ choices thus the number of such groups is
$$3cdot 4cdot 2cdot 3=72.$$
answered Jan 6 at 16:46
model_checkermodel_checker
4,18621931
answered Jan 6 at 16:46
model_checkermodel_checker
4,18621931
answered Jan 6 at 16:46
model_checkermodel_checker
4,18621931
answered Jan 6 at 16:46
model_checkermodel_checker
4,18621931
4,18621931
add a comment |
add a comment |
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