Difference between revisions of "1996 AHSME Problems/Problem 10"
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==Problem== | ==Problem== | ||
| − | How many line segments have both their endpoints located at the vertices of a given cube? | + | How many line segments have both their endpoints located at the vertices of a given [[cube]]? |
<math> \text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56 </math> | <math> \text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56 </math> | ||
| − | ==Solution 1== | + | __TOC__ |
| + | ===Solution 1=== | ||
There are <math>8</math> choices for the first endpoint of the line segment, and <math>7</math> choices for the second endpoint, giving a total of <math>8\cdot 7 = 56</math> segments. However, both <math>\overline{AB}</math> and <math>\overline{BA}</math> were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really <math>\frac{56}{2} = 28</math> line segments, and the answer is <math>\boxed{D}</math>. | There are <math>8</math> choices for the first endpoint of the line segment, and <math>7</math> choices for the second endpoint, giving a total of <math>8\cdot 7 = 56</math> segments. However, both <math>\overline{AB}</math> and <math>\overline{BA}</math> were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really <math>\frac{56}{2} = 28</math> line segments, and the answer is <math>\boxed{D}</math>. | ||
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In shorthand notation, we're choosing <math>2</math> endpoints from a set of <math>8</math> endpoints, and the answer is <math>\binom{8}{2} = \frac{8!}{6!2!} = 28</math>. | In shorthand notation, we're choosing <math>2</math> endpoints from a set of <math>8</math> endpoints, and the answer is <math>\binom{8}{2} = \frac{8!}{6!2!} = 28</math>. | ||
| − | ==Solution 2== | + | ===Solution 2=== |
Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal. | Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal. | ||
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==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=9|num-a=11}} | {{AHSME box|year=1996|num-b=9|num-a=11}} | ||
| + | |||
| + | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 07:42, 25 February 2016
Problem
How many line segments have both their endpoints located at the vertices of a given cube?
Contents
Solution 1
There are 
choices for the first endpoint of the line segment, and 
choices for the second endpoint, giving a total of 
segments. However, both 
and 
were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really 
line segments, and the answer is 
.
In shorthand notation, we're choosing 
endpoints from a set of endpoints, and the answer is 
.
Solution 2
Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.
A cube has 
edges: Four on the top face, four on the bottom face, and four that connect the top face to the bottom face.
A cube has 
square faces, each of which has facial diagonals, for a total of 
.
A cube has 
spacial diagonals: each diagonal goes from one of the bottom vertices to the "opposite" top vertex.
Thus, there are 
segments, and the answer is .
See also
| 1996 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
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