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Please clarify my doubt

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Please clarify my doubt

by dddanny2006 » Thu Jan 16, 2014 11:15 am
How many 2 element subsets can you make from the from the set {1,2,3,4,5,6}

Answer is 15,although I think it should have been 18

6 * 6 =18
2!


Its not mentioned whether the elements can be repeated or not.So I assume they can be repeated.1.What do we do with problems where order doesnt matter and they can be repeated too.
2.Also when order matters and they can be repeated.
Last edited by dddanny2006 on Thu Jan 16, 2014 11:27 am, edited 1 time in total.
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by Patrick_GMATFix » Thu Jan 16, 2014 11:26 am
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}

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by dddanny2006 » Thu Jan 16, 2014 11:30 am
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}

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by Patrick_GMATFix » Thu Jan 16, 2014 12:03 pm
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.
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by dddanny2006 » Thu Jan 16, 2014 12:16 pm
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.
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