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11 Nov 2006, 01:28
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There is a polygon with 10 sides. A point is there in the middle of the polygon and all the vertices are joined with the point and therefore producing 10 triangles. In exactly how many ways can you select three triangles no two of which are adjacent?
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11 Nov 2006, 22:12
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i have a mission !!!!!!
that is to promote the use of simple plain counting.
80 is the WRONG answer....
let's count...
for the first triangle we have 10. great.
for the second triangle we have essentially 7 (10 minus the one already chose minus the 2 adjacent). that's almost great, because there is a catch. but suppose you don't see the catch yet...
how many possibilities for the third triangle?
you think it is 4? no... that's not true... there is a catch there.
if the second triangle was nearly adjacent to the first, then we rule out only 2 triangles, not 3. and if it is not, then we rule out 3.
how to deal with this catch:
let's divide into two cases(as they say... divide and conquer):
case 1) the second triangle is nearly adjacent to the first
case 2) the second triangle is far away from the first
for case 1: we can choose the second triangle in 2 ways only. and the third traingles in 5 ways (10-3-2). so case 1 gives us: 10*2*5 = 100
for case 2: we can shoose the second triangle in 5 possible ways and the third in 4 (10-3-3). this gives us 200.
now add the two independent cases we get 300
finished? not yet, we inherently counted several options more than once, as the order of the triangles is not relevant. so we need to divide by the number of various ordering of 3 elements which is 6.
300/6 = 50.
that's the right answer.
to be more general (and you can find this in books) here are the basic rules of counting.
math rules:
"addition rule": if you have two independent cases that either one or the other can hold (but not both), add the number of possibilities.
"multiplication rule": if you have two independent cases whose relation is something like 'for each option in case 1 we can have an option of case 2' - then multiply the possibilities.
these rules are basically common sense if you apply them cautiously.
there are 2 more rules that are not math.
1) represent your counting problem in a way that it is easy to count: break it to independent stages, put things in a line or circle, understand what you are counting.
2) make sure you don't count the same thing more than once. the common mistake is repetition or ordering - if a combination is 'order irrelevant' but you counted it using the method "let's choose the first...let's choose the second..." then you must divide your result by the number of possible orderings, as you counted each possible ordering once.
3) make sure yu don't miss any possibility. if you followed rule 1, it should be fairly simple to verify.
hope it helps...
amit.
that is to promote the use of simple plain counting.
80 is the WRONG answer....
let's count...
for the first triangle we have 10. great.
for the second triangle we have essentially 7 (10 minus the one already chose minus the 2 adjacent). that's almost great, because there is a catch. but suppose you don't see the catch yet...
how many possibilities for the third triangle?
you think it is 4? no... that's not true... there is a catch there.
if the second triangle was nearly adjacent to the first, then we rule out only 2 triangles, not 3. and if it is not, then we rule out 3.
how to deal with this catch:
let's divide into two cases(as they say... divide and conquer):
case 1) the second triangle is nearly adjacent to the first
case 2) the second triangle is far away from the first
for case 1: we can choose the second triangle in 2 ways only. and the third traingles in 5 ways (10-3-2). so case 1 gives us: 10*2*5 = 100
for case 2: we can shoose the second triangle in 5 possible ways and the third in 4 (10-3-3). this gives us 200.
now add the two independent cases we get 300
finished? not yet, we inherently counted several options more than once, as the order of the triangles is not relevant. so we need to divide by the number of various ordering of 3 elements which is 6.
300/6 = 50.
that's the right answer.
to be more general (and you can find this in books) here are the basic rules of counting.
math rules:
"addition rule": if you have two independent cases that either one or the other can hold (but not both), add the number of possibilities.
"multiplication rule": if you have two independent cases whose relation is something like 'for each option in case 1 we can have an option of case 2' - then multiply the possibilities.
these rules are basically common sense if you apply them cautiously.
there are 2 more rules that are not math.
1) represent your counting problem in a way that it is easy to count: break it to independent stages, put things in a line or circle, understand what you are counting.
2) make sure you don't count the same thing more than once. the common mistake is repetition or ordering - if a combination is 'order irrelevant' but you counted it using the method "let's choose the first...let's choose the second..." then you must divide your result by the number of possible orderings, as you counted each possible ordering once.
3) make sure yu don't miss any possibility. if you followed rule 1, it should be fairly simple to verify.
hope it helps...
amit.
11 Nov 2006, 22:31
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B.T.W
I'd guess (i'm not an expert though), that the above question won't appear in gmat, or if it appears you should be happy as it says you got to the most difficult questions implying that your score is to be 50+
this is difficult. usually given as an exercise in college level math courses.
don't feel bad if you think it is too difficult to solve in 2 minutes...
be focused on the way of thinking and the principles... apply them to regular, sane, gmat questions.
amit.
I'd guess (i'm not an expert though), that the above question won't appear in gmat, or if it appears you should be happy as it says you got to the most difficult questions implying that your score is to be 50+
this is difficult. usually given as an exercise in college level math courses.
don't feel bad if you think it is too difficult to solve in 2 minutes...
be focused on the way of thinking and the principles... apply them to regular, sane, gmat questions.
amit.
