Question 58: When you draw two balls from the bin, the order does not matter here because we will consider a selection of 3 and 5 to be one set. If we were placing these balls in slots where it mattered which one went to which slot, then we will have to consider permutations.

Also, in this particular problem for calculating the probability it does not matter if you take ordered sets, you will get the same answer. For example, there will be 20 ordered sets of two balls, and then there will be 4 ordered sets of balls that give a sum of 7. So we will end up with the same answer.

On question 47, you are absolutely correct! That allows you to solve this question with zero work. I honestly didn’t even notice that, but that is a great way to do this problem and your logic is 100% solid. Personally, I don’t go in that direction because in general the harder problems will never leave a loophole like this, and when I talk about hard problems I am referring to the ones from the American Mathematics Competitions, which are very well written. But that is not the case with ACT questions, as this above example shows. It is great that you can see this and I absolutely encourage you to be on the lookout for such possibilities.

Prabh Dhaliwal says

Why would you use combination formula to determine total number of sets. Why does order not matter in this scenario?

Dabral says

Question 58: When you draw two balls from the bin, the order does not matter here because we will consider a selection of 3 and 5 to be one set. If we were placing these balls in slots where it mattered which one went to which slot, then we will have to consider permutations.

Also, in this particular problem for calculating the probability it does not matter if you take ordered sets, you will get the same answer. For example, there will be 20 ordered sets of two balls, and then there will be 4 ordered sets of balls that give a sum of 7. So we will end up with the same answer.

I hope this makes sense.

Prabh Dhaliwal says

For 47: Couldn’t you just see that Triangle ADC’s hypotenuse is 5 and that CD must be less than 5? Only option less than 5 is A. (60/13)

Dabral says

Hi Prabh,

On question 47, you are absolutely correct! That allows you to solve this question with zero work. I honestly didn’t even notice that, but that is a great way to do this problem and your logic is 100% solid. Personally, I don’t go in that direction because in general the harder problems will never leave a loophole like this, and when I talk about hard problems I am referring to the ones from the American Mathematics Competitions, which are very well written. But that is not the case with ACT questions, as this above example shows. It is great that you can see this and I absolutely encourage you to be on the lookout for such possibilities.

Cheers,

Dabral