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Exercise 6.4 · Q3

Q.How many chords can be drawn through 21 points on a circle?

Puducherry CbseNCERTSubjective· 2mImportance★★★★★est
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✓ Free question

The number of chords through 21 points on a circle is the number of ways to choose any 2 distinct points, since each chord is uniquely defined by its two endpoints. The answer is (212)=210\binom{21}{2} = 210.

The key idea here is that a chord is simply a straight line segment joining two points on the circle. Unlike a line in a plane, a chord is completely determined by its two endpoints — there is no ambiguity about which chord we mean once we pick the two points.

Why does this matter? Because the problem is not about drawing every possible line through the points (some of which might coincide or be tangents). It is about counting distinct chords. And since no three of the 21 points are collinear (they all lie on the circle), every pair of points gives a unique chord, and every chord corresponds to exactly one pair of points.

So the question reduces to: In how many ways can we select 2 distinct points from 21?

That is a pure combinations problem — order does not matter (the chord from point A to point B is the same as from B to A).

  1. Identify the total number of points: n=21n = 21.

  2. Identify the number of points needed to define one chord: r=2r = 2.

  3. Apply the combinations formula:

    The number of ways to choose rr items from nn without regard to order is

(nr)=n!r!(n−r)!.\binom{n}{r} = \frac{n!}{r!(n-r)!}.

  1. Substitute the values:

(212)=21!2!⋅19!=21×202×1.\binom{21}{2} = \frac{21!}{2! \cdot 19!} = \frac{21 \times 20}{2 \times 1}.

  1. Simplify:

21×202=21×10=210.\frac{21 \times 20}{2} = 21 \times 10 = 210.

Watch out

A common mistake is to treat this as a permutations problem and write 21×20=42021 \times 20 = 420, forgetting that the chord AB is the same as BA. Always check: does order matter? For chords, it does not.

Tip

If you ever forget the formula, think of it this way: the first point can be any of the 21, the second any of the remaining 20 — that gives 21×2021 \times 20 ordered pairs. Since each chord is counted twice (once as AB, once as BA), divide by 2: 21×202=210\frac{21 \times 20}{2} = 210.

✓Final answer

The number of chords is 210\boxed{210}.

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