A combination \(C^n_r\) counts the number of ways to select \(r\) objects from \(n\), without considering order.
A permutation \(P^n_r\) includes all possible arrangements of the selected \(r\) objects.
Therefore, multiplying \(C^n_r\) by \(r!\), the number of ways to arrange \(r\) objects, gives \(P^n_r\).

Theorem: Pascal’s Identity

Theorem:

For integers \(n\) and \(r\), where \(0 \leq r \leq n\),

\[
C^n_r + C^n_{r-1} = C^{n+1}_r
\]

Explanation:

Pascal’s Identity describes the relationship between combinations of different levels.
The left-hand side represents:
- \(C^n_r\): Choosing \(r\) items from \(n\).
- \(C^n_{r-1}\): Choosing \(r-1\) items from \(n\), then adding one more item.
The right-hand side \(C^{n+1}_r\) represents choosing \(r\) items from \(n+1\), accounting for all cases.

Example: Solving a Combination Equation

Problem:

If \(C^n_9 = C^n_8\), find \(C^{17}_n\).

Solution:

Using the combination formula:
\[
C^n_9 = \frac{n!}{9!(n-9)!}, \quad C^n_8 = \frac{n!}{8!(n-8)!}
\]
Equating the two:
\[
\frac{n!}{9!(n-9)!} = \frac{n!}{8!(n-8)!}
\]
Simplify:
\[
\frac{1}{9} = \frac{n-8}{1} \implies n = 17
\]
Therefore:
\[
C^{17}_n = C^{17}_{17} = 1
\]

Problem:

A committee of 3 persons is to be formed from 2 men and 3 women.

Solution:

Total ways to select 3 people from 5:
\[
C^5_3 = \frac{5!}{3! \cdot 2!} = 10
\]
Ways to select 1 man and 2 women:
\[
C^2_1 \cdot C^3_2 = \frac{2!}{1! \cdot 1!} \cdot \frac{3!}{2! \cdot 1!} = 2 \cdot 3 = 6
\]

Problem:

What is the number of ways to choose 4 cards from a deck of 52 cards? Solve for:

Solution:

Total ways:
\[
C^{52}_4 = \frac{52!}{4! \cdot 48!} = 270725
\]
Four cards of the same suit:
\[
4 \cdot C^{13}_4 = 4 \cdot \frac{13!}{4! \cdot 9!} = 4 \cdot 2860 = 11440
\]
Four cards from different suits:
\[
C^{13}_1 \cdot C^{13}_1 \cdot C^{13}_1 \cdot C^{13}_1 = 13^4 = 28561
\]

Four face cards:
\[
C^{12}_4 = \frac{12!}{4! \cdot 8!} = 495
\]