<p>In how many ways can 10 people be divided into two grojups, one with 7 people and the other with 3 people?</p>
<p>Ans. 120</p>
<p>Please explain this to me (y u get 120 as an answer)</p>
<p>Thanks in advance</p>
<p>OK, what you are talking about is combinations. You need to know how many combinations of three you can get out of your ten people. If you get your combinations of three sorted out, then the people left over will automatically fall into a group of seven, so we'll just think about the three (because its easier).</p>
<p>If I take three ppl from a group of ten, then there are (10 x 9 x 8) ways to do it. Now, that takes into account the fact that we chose one particular person to be 1st, one to be 2nd, and one to be 3rd. But in your problem we don't care about who's first, etc., we just want a group. So we have to divide by (3!) to get rid of the effect of the order in (10 x 9 x 8). Thats because if we choose the ppl in order we will get more possibilities than if we just have them in a group. So the answer is (10 x 9 x 8)/(3 x 2 x 1) = 120. The more general formula for combinations is n!/[(n-r)!r!] where n is the total number of items, and you want groups taken r at a time. I presume your are familiar with factorial symbol. Otherwise you can just use the nCr function on your calculator.</p>
<p>Hope that helps, and just ask if you need a better explanation.</p>
<p>*** this one's also tricky. So we have 10 people right? We want 7 people in one group, and 3 in another. How many combinations? To make things simpler, let us have this assumption: If we have set who the 7 people in one group are, we will have automatically set who the 3 people in the other group is. To simplify things, let's consider making the group of 3 instead of the group of 7. SO, okay, we have 10 people. If we want to make a group of 3 out of this 10, we have 10!/(10-3)! = 10<em>9</em>8 combinations. Thus, there are 10<em>9</em>8 total possible combinations (I am using the term combination loosely) of groups of 7 and 3. BUT notice that if, say, we have people A, B, C in the 3-member group, this combination itself accounts for 6 combinations already. But what we want, of course, is the number of groups possible. The people A, B, C whould make only 1 group, not 6. Thus, we divide 10<em>9</em>8 by 6, and get 120 groups.</p>
<p>you do 10 computation3 on your calculator</p>