A convex polygon has 44 diagonals. Find the number of its sides.

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Concept understanding — Combinations Selection

Combinations: Choosing Without Ordering

Imagine you're picking a team of 3 players from a group of 5 friends: Alice, Bob, Charlie, Deepa, and Esha. The team {Alice, Bob, Charlie} is the same team as {Bob, Charlie, Alice} — the order you name them doesn't matter. What matters is which 3 people you pick.

That's the core idea of combinations: selection without regard to order.

The Intuition: Why Order Doesn't Matter

Let's contrast with permutations. If you were assigning positions — captain, vice-captain, treasurer — then {Alice as captain, Bob as vice-captain, Charlie as treasurer} is different from {Bob as captain, Alice as vice-captain, Charlie as treasurer}. Order matters there.

But for a plain team, a committee, a hand of cards, or a set of toppings on a pizza — order is irrelevant. You just care about which items are chosen.

Note

Key distinction: Permutations count arrangements (order matters). Combinations count selections (order doesn't matter).

From Permutations to Combinations

Suppose you want to choose 2 letters from {A, B, C}. If order mattered, you'd have these 6 permutations:

AB, BA, AC, CA, BC, CB

But if order doesn't matter, AB and BA are the same selection. So the distinct combinations are just:

{A, B}, {A, C}, {B, C} — only 3.

Notice the pattern: each combination of 2 items corresponds to $2! = 2$ permutations (because you can arrange those 2 items in 2 ways). So:

\text{Number of combinations} = \frac{\text{Number of permutations}}{r!}

Where $r$ is the number of items you're choosing.

The Precise Statement

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

This is read as "n choose r" and gives the number of ways to select $r$ distinct objects from a set of $n$ distinct objects, where order does not matter.

Conditions:

$n$ and $r$ are non-negative integers
$r \leq n$
The objects are distinct (no repetitions)

Why the Formula Works

Start with permutations of $r$ items from $n$ : $P(n, r) = \frac{n!}{(n-r)!}$ .

Each combination of $r$ items can be arranged in $r!$ different orders. So the number of combinations is the number of permutations divided by the number of ways to rearrange each selection:

\binom{n}{r} = \frac{P(n, r)}{r!} = \frac{n!}{r!(n-r)!}

Tip

A quick check: $\binom{n}{0} = 1$ (there's exactly one way to choose nothing), and $\binom{n}{n} = 1$ (one way to choose everything).

A Concrete Example

How many different 5-card hands can be dealt from a standard 52-card deck?

Here, $n = 52$ , $r = 5$ . The hand {A♠, K♥, Q♦, J♣, 10♠} is the same regardless of the order you receive the cards.

\binom{52}{5} = \frac{52!}{5! \cdot 47!} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2,598,960

That's over 2.5 million possible hands — which is why poker is interesting. …

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The number of diagonals in a convex polygon with $n$ sides is given by the formula $D = \frac{n(n-3)}{2}$ . By setting $D=44$ and solving the resulting quadratic equation, we find that the polygon has 11 sides.

Concept and Intuition

A convex polygon is a closed shape with straight sides where all interior angles are less than $180^\circ$ . A diagonal of a polygon is a line segment connecting two non-adjacent vertices. The sides of the polygon connect adjacent vertices.

To understand how to count diagonals, let's think about the total number of ways to connect any two vertices in a polygon.

If a polygon has $n$ vertices, and we pick any one vertex, we can draw a line segment from it to any of the other $n-1$ vertices.

If we do this for all $n$ vertices, we would get $n(n-1)$ line segments. However, this counts each segment twice (e.g., the segment from vertex A to vertex B is counted once when we consider vertex A, and again when we consider vertex B). So, the total number of unique line segments connecting any two vertices is $\frac{n(n-1)}{2}$ .

These $\frac{n(n-1)}{2}$ segments include both the sides of the polygon and its diagonals. A polygon with $n$ vertices also has $n$ sides.

Therefore, to find the number of diagonals, we subtract the number of sides from the total number of segments:

Number of diagonals = (Total number of segments connecting any two vertices) - (Number of sides)

Number of diagonals = $\frac{n(n-1)}{2} - n$

Let's simplify this expression:

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

D

This formula is fundamental for problems involving diagonals of polygons.

Step-by-step Solution

Identify the given information:

We are given that the convex polygon has $44$ diagonals. So, $D = 44$ .
Apply the formula for the number of diagonals:

We use the formula derived above:

D = \frac{n(n-3)}{2}

Substitute the given value of $D$:

44 = \frac{n(n-3)}{2}

Formulate a quadratic equation: Multiply both sides by $2$ :

44 \times 2 = n(n-3)

88 = n^2 - 3n

Rearrange the terms to form a standard quadratic equation $an^2 + bn + c = 0$:

n^2 - 3n - 88 = 0

Solve the quadratic equation for $n$ : We can solve this quadratic equation by factoring or by using the quadratic formula. …

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Q.A convex polygon has 444444 diagonals. Find the number of its sides.

Concept and Intuition

Step-by-step Solution

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Q.A convex polygon has $44$ diagonals. Find the number of its sides.

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