Please clarify my doubt

If we are asked to make subsets, the elements cannot be repeated. {1,1} is not a subset of {1,2,3,4,5,6}. It would be a subset of {1,1,2,3,4,5,6}

-Patrick

Thanks for that Patrick.What if we have a situation where-
1.Order doesnt matter and they can be repeated too.
2. Order matters and they can be repeated.

Patrick_GMATFix wrote:If we are asked to make subsets, the elements cannot be repeated. {1,1} is not a subset of {1,2,3,4,5,6}. It would be a subset of {1,1,2,3,4,5,6}

-Patrick

1.Order doesn't matter and they can be repeated too.
Since digits can be repeated, we pick from 6, then from 6. That's 36 pairs. We need to remove duplicates (such as {1,2} and {2,1} since order doesn't matter. 6 of the 36 pairs are the same digit repeated ({1,1} to {6,6}) and do not suffer from duplication. The other 30 pairs are really 15 pairs, each duplicated. So the number of possible selections under this scenario is 6 + 15 = 21 pairs.

The problem with the solution you came up with (6*6/2 = 18) is that it assumes that every one of the 36 pairs has a duplicate. In fact, only pairs of different digits have a duplicate. {1,1} and {1,1} are one and the same they're the result of "pick 1, then pick 1". To make sense of this, imagine a much smaller universe where we have to pick a pair from {1,2} in which order doesn't matter and repeats are allowed. Your formula would yield (2*2/2 = 2 pairs). The real answer is 3 pairs possible {1,1}, {1,2} and {2,2}. Only pairs with different digits have duplicates.

2. Order matters and they can be repeated.
This is the same as the problem above, but we do not need to remove duplicates. This is basically the same as how many outcomes are possible when two six-sided dice are thrown. There are 6 * 6 or 36 pairs possible.

Thats a brilliant explanation.Thank you,I guess there's no formula that could directly get you the number of pairs.We have to rely on normal counting.

Patrick_GMATFix wrote:1.Order doesn't matter and they can be repeated too.
Since digits can be repeated, we pick from 6, then from 6. That's 36 pairs. We need to remove duplicates (such as {1,2} and {2,1} since order doesn't matter. 6 of the 36 pairs are the same digit repeated ({1,1} to {6,6}) and do not suffer from duplication. The other 30 pairs are really 15 pairs, each duplicated. So the number of possible selections under this scenario is 6 + 15 = 21 pairs.

The problem with the solution you came up with (6*6/2 = 18) is that it assumes that every one of the 36 pairs has a duplicate. In fact, only pairs of different digits have a duplicate. {1,1} and {1,1} are one and the same they're the result of "pick 1, then pick 1". To make sense of this, imagine a much smaller universe where we have to pick a pair from {1,2} in which order doesn't matter and repeats are allowed. Your formula would yield (2*2/2 = 2 pairs). The real answer is 3 pairs possible {1,1}, {1,2} and {2,2}. Only pairs with different digits have duplicates.

2. Order matters and they can be repeated.
This is the same as the problem above, but we do not need to remove duplicates. This is basically the same as how many outcomes are possible when two six-sided dice are thrown. There are 6 * 6 or 36 pairs possible.