6.3 Q-1

Exercise 6.3 — Question 1

This exercise covers problems involving combinations, including the symmetry property and applications such as polygon diagonals.

---

Key Formula

The combination formula is:

${}^n{C}_r=\frac{n!}{r!(n-r)!}$

Symmetry Property of Combinations

If ${}^n{C}_p={}^n{C}_q$, then either:

$p=q\text{or}p+q=n$

This follows from the identity ${}^n{C}_r={}^n{C}_{n-r}$.

---

Worked Examples

Example 1: Find $n$ if ${}^n{C}_{15}={}^n{C}_7$

Using the symmetry property:

$p+q=n⟹15+7=n⟹n=22$

(Since $15\ne 7$, we cannot use $p=q$.)

Answer: $n=22$

---

Example 2: Find $n$ if ${}^n{C}_3={}^n{C}_{n-3}$

By the symmetry property ${}^n{C}_r={}^n{C}_{n-r}$, this is always true for any $n\ge 3$.

If instead the problem gives ${}^n{C}_3={}^n{C}_5$, then:

$3+5=n⟹n=8$

---

Example 3: Number of Diagonals in a Polygon

An $n$-sided polygon has $n$ vertices. The total number of line segments joining any two vertices is ${}^n{C}_2$. Subtracting the $n$ sides gives the number of diagonals:

$\text{Diagonals}={}^n{C}_2-n=\frac{n(n-1)}{2}-n=\frac{n(n-3)}{2}$

Example: A hexagon ($n=6$) has $\frac{6\times 3}{2}=9$ diagonals.

---

Example 4: Conditional Selection

In how many ways can a team of 4 be chosen from 7 people if one specific person must always be included?

Since one person is fixed, we choose the remaining $4-1=3$ from the remaining $7-1=6$ people:

${}^6{C}_3=\frac{6!}{3!\cdot 3!}=20$

Answer: 20 ways

---