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Zorluk: ZorCombinations and Group Selections

In a plane, there are nn points such that no three points are collinear. If the number of distinct triangles that can be formed using these points as vertices is exactly 55 times the number of distinct line segments that can be formed by joining pairs of these points, what is the value of nn?

  1. A
    77
  2. B
    1515
  3. C
    1616
  4. 1717Cevap
  5. E
    3232

Cevap

The correct value of nn is 1717.
The number of distinct triangles formed by nn points (no three collinear) is (n3)=n(n−1)(n−2)6\binom{n}{3} = \frac{n(n-1)(n-2)}{6}, and the number of line segments is (n2)=n(n−1)2\binom{n}{2} = \frac{n(n-1)}{2}. Setting (n3)=5(n2)\binom{n}{3} = 5 \binom{n}{2} gives n(n−1)(n−2)6=5n(n−1)2\frac{n(n-1)(n-2)}{6} = \frac{5n(n-1)}{2}. Dividing both sides by n(n−1)2\frac{n(n-1)}{2} simplifies the equation to n−23=5\frac{n-2}{3} = 5, yielding n−2=15n - 2 = 15, so n=17n = 17.

Adım Adım Çözüm

1
Express the number of triangles and line segments using combinations.
The number of triangles formed by selecting 33 non-collinear points from nn points is (n3)=n(n−1)(n−2)6\binom{n}{3} = \frac{n(n-1)(n-2)}{6}. The number of line segments formed by selecting 22 points from nn points is (n2)=n(n−1)2\binom{n}{2} = \frac{n(n-1)}{2}.
Order does not matter when selecting vertices for triangles or endpoints for line segments.
2
Set up the given equation relating the two quantities.
n(n−1)(n−2)6=5×n(n−1)2\frac{n(n-1)(n-2)}{6} = 5 \times \frac{n(n-1)}{2}
The problem states that the number of triangles is 55 times the number of line segments.
3
Simplify the equation for n≥3n \ge 3.
Divide both sides by n(n−1)2\frac{n(n-1)}{2} to get n−23=5\frac{n-2}{3} = 5, which simplifies to n−2=15n - 2 = 15.
Since n≥3n \ge 3, n(n−1)≠0n(n-1) \ne 0, so dividing by common terms is valid.
4
Solve for nn.
n = 15 + 2 = 17
Adding 22 to both sides isolates nn.

Anahtar Kavram

Combinations in Geometry
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Konular

Combinations and Group Selections

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