In a plane, there are n points such that no three points are collinear. If the number of d… — GMAT

In a plane, there are $n$ 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 $5$ times the number of distinct line segments that can be formed by joining pairs of these points, what is the value of $n$ ?

A
$7$
B
$15$
C
$16$
$17$ Answer
E
$32$

Answer

The correct value of

n

17

The number of distinct triangles formed by

n

points (no three collinear) is

\binom{n}{3} = \frac{n(n-1)(n-2)}{6}

, and the number of line segments is

\binom{n}{2} = \frac{n(n-1)}{2}

. Setting

\binom{n}{3} = 5 \binom{n}{2}

gives

\frac{n(n-1)(n-2)}{6} = \frac{5n(n-1)}{2}

. Dividing both sides by

\frac{n(n-1)}{2}

simplifies the equation to

\frac{n-2}{3} = 5

, yielding

n - 2 = 15

, so

n = 17

Step-by-Step Solution

Express the number of triangles and line segments using combinations.

The number of triangles formed by selecting

3

non-collinear points from

n

points is

\binom{n}{3} = \frac{n(n-1)(n-2)}{6}

. The number of line segments formed by selecting

2

points from

n

points is

\binom{n}{2} = \frac{n(n-1)}{2}

Order does not matter when selecting vertices for triangles or endpoints for line segments.

Set up the given equation relating the two quantities.

\frac{n(n-1)(n-2)}{6} = 5 \times \frac{n(n-1)}{2}

The problem states that the number of triangles is

5

times the number of line segments.

Simplify the equation for

n \ge 3

Divide both sides by

\frac{n(n-1)}{2}

to get

\frac{n-2}{3} = 5

, which simplifies to

n - 2 = 15

Since

n \ge 3

n(n-1) \ne 0

, so dividing by common terms is valid.

Solve for

n

n = 15 + 2 = 17

Adding

2

to both sides isolates

n

Key Concept

Combinations in Geometry

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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