<p>But how did you get (2*1). I thought it would be 3!/ 3!(3-1)! ?</p>
<p>3!/(3!)(3-1)! would be a fraction, dude. The combination formula is nCr = n!/[(n-r)!*(r!)]</p>
<p>I just punch them into my TI-84.... it has a permutation and combination solver...</p>
<p>So Arachnotron agree with me?</p>
<p>No belly you just wrote it wrong or something. It's what Arachnotron wrote. The idea is to consider all the permutations and then "collapse" them by dividing out. So if you want all the permutations of 5 elements you have </p>
<p>ABCDE</p>
<p>which is 5! because you have 5 choices for the first slot, 4 for the second, 3 for the third, 2 for the fourth, 1 for the last slot. (That alone deserves a post for justification.) But if you want some of them to be the same, you basically divide out by all the ways you would have been able to order them. </p>
<p>So if you wanted to know how many ways you could make a three-person basketball team from five potential players, the order of the choices for the three players does not matter. The team ABC = the team BAC. There are 3!=6 ways to order these elements, so if you counted them like that you would have been counting 6 distinct orders when you only want to count 1 of them. Dividing out by 6 fixes this. </p>
<p>But, and this is kind of subtle, you have to factor in the fact that the order of the players who are not chosen doesn't matter, either. The unchosen players DE = the unchosen players ED. There are two ways to order these two elements, so if you counted them like that you would have been counting 2 distinct orders when you only want to count 1 of them. Dividing out by 2 fixes this. In general, to divide out the n elements you just find the number of ways to order them. So in these two cases you divide out by 3!*2!. </p>
<p>Essentially you make three of the elements the same, and two of the elements the same too but different from the first one. You started with</p>
<p>ABCDE</p>
<p>but manipulated it so you effectively are counting all the permutations of </p>
<p>AAABB</p>
<p>The visual way of thinking about it like this works because AAA = AAA no matter how much you swap them. Take AAA and swap the first two, you still get AAA. This is not true if you swap A and B in ABC because you would get BAC, which is distinct but you want it to be the same. </p>
<p>The advantage of thinking about it this way is that you can "collapse" by as many groups as you want. If you have 9 people and you want all the ways you could put them in three hotel rooms that hold three people each, you do the same procedure but collapse three groups of three. So</p>
<p>ABCDEFGHI</p>
<p>becomes</p>
<p>AAABBBCCC </p>
<p>= 9!/(3!<em>3!</em>3!)</p>
<p>The choose formula does not account for this. In general you just divide out by the factorial of however many elements are in each group, and you just have to make sure that they all add up to the total number of elements (i.e. 3+2=5 in the basketball case, 3+3+3=9 in the hotel room case). Does that help?</p>
<p>Thank you all for reply at length. I think about it today and I found out that I've been using the wrong formula all along ( I messed combination and permutation up xD). Just a curious answer: What do you have if you plus 3+4+5 instead of multiplying?</p>
<p>OK I take it the question is, what question would you be answering if you answered 3+4+5 in the context of the original problem, which was</p>
<p>
[quote]
You can choose 1 main disk, 1 drink, 1 dessert from 3 main dishes, 4 drinks and 5 desserts. How many combination you can get?
[/quote]
</p>
<p>The answer 3+4+5 would be correct for the question:</p>
<p>If you have 3 main dishes, 4 drinks, and 5 desserts, and can only choose exactly ONE of main dish, drink, OR dessert, how many possible (impoverished) meals could you have?"</p>
<p>In this case, you could have either one of 3 main dishes, or one of 4 drinks, or one of 5 desserts, and they do not <em>combine</em> to make a full meal.</p>