6.3 Q-5 | Permutation, Combination And Probability | Mathematics 11

Exercise 6.3 — Question 5

Solve for $n$ if $^{n} C_{15} =^{n} C_{7}$ .

For combinations, the following symmetry property holds:

$^{n} C_{p} =^{n} C_{q} ⟹ p = q or p + q = n$

This follows directly from the formula:

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

Since $^{n} C_{r} =^{n} C_{n - r}$ , if $^{n} C_{p} =^{n} C_{q}$ then either $p = q$ or $q = n - p$ , i.e., $p + q = n$ .

Given: $^{n} C_{15} =^{n} C_{7}$

Applying the symmetry property:

Case 1: $15 = 7$ — This is impossible.

Case 2: $15 + 7 = n$

$n = 22$

Verification:

$^{22} C_{15} =^{22} C_{7} = \frac{22 !}{7 ! \cdot 15 !} ✓$

$n = 22$

Condition	Result
$^{n} C_{p} =^{n} C_{q}$ and $p = q$	Trivially true
$^{n} C_{p} =^{n} C_{q}$ and $p \neq = q$	$n = p + q$

Solve for $n$ if $^{n} C_{15} =^{n} C_{7}$ .

For combinations, the following symmetry property holds:

$^{n} C_{p} =^{n} C_{q} ⟹ p = q or p + q = n$

This follows directly from the formula:

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

Since $^{n} C_{r} =^{n} C_{n - r}$ , if $^{n} C_{p} =^{n} C_{q}$ then either $p = q$ or $q = n - p$ , i.e., $p + q = n$ .

Given: $^{n} C_{15} =^{n} C_{7}$

Applying the symmetry property:

Case 1: $15 = 7$ — This is impossible.

Case 2: $15 + 7 = n$

$n = 22$

Verification:

$^{22} C_{15} =^{22} C_{7} = \frac{22 !}{7 ! \cdot 15 !} ✓$

$n = 22$

Condition	Result
$^{n} C_{p} =^{n} C_{q}$ and $p = q$	Trivially true
$^{n} C_{p} =^{n} C_{q}$ and $p \neq = q$	$n = p + q$