<p>Philip is on a committee that has 6 members. How many different subcommittees of 3 can he forme that have philip in them?
a)3
b)6
c)10
d)15
e)20</p>
<p>what is the greatest number of balls of radius 1 in that will fit into a covered wooden box with interior dimensions of 2in by 4in by 4 in?</p>
<p>a)2
b)4
c)8
d)16
e) 32</p>
<p>ten distinct lines lie in the same plane. No pair of these lines is parallel o each other, and no more than 2 of these lines intersect at any one point. How many points lie on more than one of these lines?</p>
<ol>
<li>Consider a group that contains phillip, x, and y. There are 5 possibilities for x, since phillip is removed from the group of 6. There are 4 possibilities for y, since phillip and x are removed from the 6. Thus, there are 4x5 possibilities, or 20. C</li>
</ol>
<p>You can also just do 6 choose 3, which is 6!/(3!*3!)=20</p>
<ol>
<li>Consider a box that is 4 in wide, 4 in long, and 2in high. It is evident that only 4 balls at the maximum can fit on the bottom. This is due to the diameter of the balls- each has a diameter of 2, so two of them in a row take up the whole width and two in a row take up the whole length. One ball is also 2 in tall, so there is only one layer in this scenario. Since there can only be 2x2 balls, the answer is 4. B</li>
</ol>
<p>(I am somewhat unsure of that one).</p>
<ol>
<li>The first line intersects with 9 other lines since the lines are all non-parallel and on the same plane. The second line intersects with 8 other lines (we exclude the first line), the third line intersects with 7 other lines, etc. The answer is 9+8+7…, or 45. C</li>
</ol>
<p>I could be wrong on all of these, but I hope not!</p>
<p>Ummm… I agree with ur #1 and 2… Although It might be 5C2= 10 in #1 (but i’m not convinced)</p>
<h1>3 tho… I can’t make heads or tales of it. I can’t picture them at all the way u described them. My answer would have been maybe they formed a decagon, but again, a decagon had parallel lines x.X</h1>
<h1>1 is, indeed 10, as 5C2 is the proper expression. As order is not important (ABC and ACB would be considered the same committee), the 20 possible permutations would only be 10 possible combinations.</h1>
<p>Okay, since each line is not parallel, all lines are on the same plane, and all intersections are of two lines, each line intersects with each other line. Each of the ten lines has 9 intersections. However, we cannot simply do 9x10 because that would count some intersections twice. </p>
<p>So, instead, we know that line 1 will intersect with lines 2,3,4…10. So it has 9 intersections. Line 2 also has 9 intersections, but we have already counted its intersection with line 1, so we give it 8. Line 3 has 9 intersctions, but we have already counted two of them, so we give it 7 intersections. We continue this until the 10th line. We have already counted all of its intersections. So we evaluate 9+8+7+6+5+4+3+2+1=45.</p>
<p>I am very stupid.
Now I know that thank you for enlightening me :P</p>
<p>I’m sorry your struggling with me… but last request… maybe, just maybe, it might help me understand if u draw them and send me a pic or snapshot or sth?</p>
<p>what’s confusing me is that the question says “NO MORE THAN 2 OF THESE LINES INTERSECT AT ANY ONE POINT”
and you’re saying “we know that line 1 will intersect with lines 2,3,4…10. so…”</p>
<p>Not this. 6C3 would include subcommittees that don’t include Philip. “Consider a group that contains phillip, x, and y” was exactly right, but it should lead you to doing 5C2, not 5P2.</p>
<p>What “no more than 2 of these lines intersect at any one point” means is that there is no point where three lines or more come together. My calculation reflects that. There are still 45 places where two lines intersect. I don’t know how else to explain it without using a visual.</p>
<p>Also, 6c3 was wrong, I get it :). Doing the P, x, y scenario would have worked if I divided the total combos by 2 to eliminate the scenarios where x and y are simply swapped.</p>