<p>2) NOW, how many groups of 2 can be made outta 3 ppl?</p>
<p>1, 2
1, 3
2, 3</p>
<p>and the answer 3. are these questions asking different things here? like in tha first question, would assigning guy 1 and guy 2 in that order be one way, and would assigning guy 2 first and guy 1 second be another way?</p>
<p>It IS the same question both times. But only your second answer is correct. The reason your first answer is wrong is that you have counted every group twice. Yes you have 3 choices for the first kid and then 2 for the second kid. But that would count AB and BA as two different groups…</p>
<p>BTW, there is another easier way to answer this particular question: To make groups of 2 out of 3 people, ask yourself how many ways are there to choose the one kid who will NOT be in the group. Obviously, there are three different kids you could pick. </p>
<p>In formal language, n C r = n C (n-r). It just means that the number of ways to pick a sub-group to be in a team is the same as the number of ways to pick the rest of the kids to not be in the team. For example, 7C2 = 7C5. This is not usually that useful, but in your problem, it helps because 3C2 = 3C1 and 3C1 is just 3.</p>
<p>alright, how about another one: at my prep class, the answer to this was 30, but after listing non-repeating combos, i found the answer to be 18…</p>
<p>joe has 6 different prints, and can display 2 of them on his wall at the same time. how many different ways can he display 2 prints?
they told me 6 * 5 = 30, but that would include repeating combos… ***?</p>
<p>and there was yet another question on here a while ago that asked, how many ways can you make a group of 3 out of 10 people? and many agreed the answer was 10 * 9 * 8 = 720… wouldnt that be counting so many repeating groups?! im still confused!</p>
<p>uh like, i’m not sure how you got 18. and yeah it would include repeating combos (like AB and BA) but in this context, those would be considered different. thus they’d be right.</p>
<p>and for the other question:
yeah, it would be counting many repeating groups. to find out how many, let’s usa a group ABC as an example. the permutations of ABC are:
ABC
ACB
BAC
BCA
CAB
CBA
so there are 6 permutations. alternatively, just note that it is 3! (3 factorial).</p>
<p>then the answer would be 720/6 (to correct for overcounting of the same groups), giving us 120. alternatively, just do 10C3, as pckeller explained</p>