<p>in how many ways can 10 people be divided into two groups, one with 7 people and the other with 3 people?!</p>
<p>lol I just did that on the math2c practice exam in the Real SAT II book...I'd like to know as well</p>
<p>Unless I'm misreading the question, this would just be a combination problem. C (10,3) which is 120?</p>
<p>yah thats right...how do u differentiate between combination and permutation problems?</p>
<p>Usually, somewhere in a combination problem, the word "combination" is mentioned, whereas in a permutation problem, the word "arrangement" is mentioned. Basically, in permutations, order matters (A, B, C would not be the same as A, C, B) while in combinations, order does not matter.</p>
<p>So combinations would generally have a smaller # than permutations?</p>
<p>isn't it
C(10,7)*C(10,3)? :S</p>
<p>No</p>
<p>C(10,7)=C(10,3)</p>
<p>no its not C(10,7)*C(10,3) because once you figure out 7 people that will be in a group, the rest 3 will automatically be in the other group.</p>
<p>You use combination when you don't need to consider the order of the things, while you use permutation if order matters. Thus, permutation will always have a greater value than combination (except of course the trivial situations such as C(1,1) P(1,1) etc.), since taking order into account creates more possibilities for arranging a set number of things.</p>
<p>An easy way to think about is, is to draw boxes, with how many numbers you can pick from on top, and if it is a combinations problem, draw the same number of boxes on the bottom, and do the largest factorial that fits in there. So 10<em>9</em>8 / 3<em>2</em>1.</p>