12.2 Permutations, Combinations, Binomial Expansion & Probability (Olasılık)

12.2 Permutations, Combinations, Binomial Expansion & Probability (Olasılık)

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1. Permutations (Permütasyon — Order Matters)

Fundamental Counting Principle & Linear Permutations

• Addition Rule (Toplama Yoluyla Sayma): If disjoint tasks $A$ and $B$ can be done in $m$ and $n$ ways respectively, choosing one task OR (veya) the other can be done in $m + n$ ways.
• Multiplication Rule (Çarpma Yoluyla Sayma): If sequential stages $A$ and $B$ can be done in $m$ and $n$ ways, performing $A$ AND (ve) $B$ together can be done in $m \cdot n$ ways.
• Linear Permutations (Sıralama): The number of ordered arrangements of $r$ items chosen from $n$ distinct items is:

$$P(n, r) = \frac{n!}{(n - r)!} = \underbrace{n(n - 1)(n - 2)\cdots(n - r + 1)}_{r \text{ factors}}, \qquad P(n, n) = n!$$

(Recall $0! = 1$ and $1! = 1$.)

Block-Glue Method (Paketleme Yöntemi) vs. Gap-Insertion Method

• Items that MUST stay together (Yan Yana Olma Koşulu): Tie the $k$ items that must be adjacent into 1 single super-block (paket). If there are $m$ other free items, first arrange the $(m + 1)$ units in $(m + 1)!$ ways, then multiply by $k!$ for the internal rearrangements inside the block:

$$\text{Total Ways} = (m + 1)! \cdot k!$$

• Items that MUST NOT be adjacent (Yan Yana Olmama Koşulu): When only 2 items must stay apart, subtract the glued cases from total arrangements: $n! - (n-1)! \cdot 2!$. When 3 or more items of one group ($B$) cannot be adjacent, do not subtract from the total! Instead, use the gap-insertion method (boşluk yöntemi): first seat the $m$ unrestricted items ($A$) in $m!$ ways, creating $m + 1$ gaps (_ A _ A _ ... _ A _), then choose and order the $k$ restricted items into those $m + 1$ gaps in $P(m + 1, k)$ ways.

Circular Permutations (Dairesel Permütasyon)

• Around a Round Table (Yuvarlak Masa): Rotating everyone by one seat produces the same relative arrangement, so fix 1 person as a reference landmark and permute the remaining $n - 1$ people:

$$P_{\text{circle}}(n) = (n - 1)!$$

• On a Flip-Symmetric Keyring (Maskotlu / Maskotsuz Anahtarlık): A plain circular keyring without a fixed clasp can be flipped over in 3D space, making clockwise and counterclockwise orders identical:

$$P_{\text{keyring}}(n) = \frac{(n - 1)!}{2} \quad (n \ge 3)$$

Permutations with Repetition (Tekrarlı Permütasyon)

The number of distinct permutations of $n$ objects where $n_1$ are identical of type 1, $n_2$ are identical of type 2, ..., and $n_k$ are identical of type $k$ ($n_1 + n_2 + \dots + n_k = n$) is:

$$\frac{n!}{n_1! \cdot n_2! \cdots n_k!}$$

• Leading-Zero Digit Rule: When forming an $n$-digit number using all $n$ given digits where the digit $0$ appears $n_0$ times, $0$ cannot occupy the leftmost place. Multiply the total repeated permutation count by the fraction of non-zero digits:

$$\text{Valid } n\text{-digit numbers} = \frac{n!}{n_1! \cdot n_2! \cdots n_k!} \cdot \frac{n - n_0}{n}$$

• Grid Shortest Paths: Walking on a rectangular grid from corner $A$ to corner $B$ using $R$ steps Right and $U$ steps Up is a repeated permutation of $R$ identical R's and $U$ identical U's: $\frac{(R + U)!}{R! \cdot U!} = \binom{R + U}{R}$.

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2. Combinations (Kombinasyon — Selection Without Order) & Binomial Theorem

Core Combination Identities

The number of ways to select an unordered subset of $r$ elements from $n$ elements is:

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

• Symmetry Rule: $\binom{n}{r} = \binom{n}{n - r}$. Consequently, if $\binom{n}{x} = \binom{n}{y}$, then either $x = y$ or $x + y = n$.
• Pascal's Identity: $\binom{n}{r} + \binom{n}{r + 1} = \binom{n + 1}{r + 1}$.
• Total Subsets Sum: $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n$.

Geometric Counting with Combinations

Given $n$ points in a plane:

• If no three points are collinear (e.g., $n$ points on a circle), they determine $\binom{n}{2}$ distinct lines and $\binom{n}{3}$ triangles. The count $\binom{n}{4}$ of convex quadrilaterals additionally requires the points to be in convex position, as with distinct points on a circle.
• If $k$ of the $n$ points are collinear on a single line $d$ (and the other $n - k$ are in general position), any 3 points chosen solely from line $d$ fail to form a triangle. Subtract the degenerate cases:

$$\text{Number of Triangles} = \binom{n}{3} - \binom{k}{3}, \qquad \text{Number of Distinct Lines} = \binom{n}{2} - \binom{k}{2} + 1$$

• Parallelograms from Parallel Lines: A grid of $m$ parallel horizontal lines intersecting $n$ parallel slanted lines forms $\binom{m}{2} \cdot \binom{n}{2}$ parallelograms.

Binomial Theorem (Binom Açılımı)

For the expansion of $(ax^p + by^q)^n$:

• There are $n+1$ indexed terms before collecting like powers; distinct nonzero terms require nondegenerate coefficients and exponents.
• Sum of coefficients (Katsayılar Toplamı): Substitute $x = 1, y = 1$ to get $(a + b)^n$.
• General $(r + 1)$-th Term Formula:

$$T_{r+1} = \binom{n}{r} (ax^p)^{n-r} (by^q)^r$$

• Constant Term (Sabit Terim) Trap: For a polynomial expression, the constant term is found by setting $x = 0$. However, in rational binomials such as $\left(2x^2 - \frac{1}{x}\right)^n$, substituting $x = 0$ causes division by zero! Instead, write the general term $T_{r+1}$, combine the powers of $x$ into $x^{p(n-r) + qr}$, and solve $p(n - r) + qr = 0$ for $r$.

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3. Probability (Olasılık)

For an experiment with sample space $E$ (örnek uzay) of equally likely outcomes and an event $A$ (olay):

$$P(A) = \frac{s(A)}{s(E)} = \frac{\text{Desired Outcomes (İstenen Durum)}}{\text{All Possible Outcomes (Tüm Durum)}}, \qquad 0 \le P(A) \le 1$$

• Complement Rule (Tümleyen): $P(A') = 1 - P(A)$. Whenever a problem asks for the probability of "at least one (en az bir)", compute $1 - P(\text{none})$.
• Union Rule (Birleşim): $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If $A$ and $B$ are mutually exclusive (ayrık olaylar), $P(A \cap B) = 0$.
• Independent Events (Bağımsız Olaylar): Two events that do not affect each other satisfy $P(A \cap B) = P(A) \cdot P(B)$.
• Conditional Probability (Koşullu Olasılık): The probability of $A$ given that $B$ has already occurred ("... bilindiğine göre") restricts the denominator to $B$:

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{s(A \cap B)}{s(B)}$$

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Summary Reference Table

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Worked Examples

Worked Example 1: Repeated Permutation with Leading Zeros & Binomial Constant Term

Problem:

1. How many distinct $6$-digit numbers can be formed using all the digits of $202440$?
2. Find the constant term (sabit terim) in the expansion of $\left(2x^2 - \frac{1}{x}\right)^6$.

Step-by-Step Solution:

1. The $6$ digits are $\{2, 0, 2, 4, 4, 0\}$: the digit $2$ repeats $2$ times, $4$ repeats $2$ times, and $0$ repeats $2$ times ($n = 6, n_0 = 2$). Using the repeated permutation formula with the leading-zero fraction $\frac{6 - 2}{6} = \frac{4}{6}$:

$$\text{Count} = \frac{6!}{2! \cdot 2! \cdot 2!} \cdot \frac{4}{6} = \frac{720}{8} \cdot \frac{2}{3} = 90 \cdot \frac{2}{3} = 60$$

1. Write the general $(r + 1)$-th term of $\left(2x^2 - x^{-1}\right)^6$:

$$T_{r+1} = \binom{6}{r} (2x^2)^{6-r} (-x^{-1})^r = \binom{6}{r} 2^{6-r} (-1)^r x^{12 - 2r - r} = \binom{6}{r} 2^{6-r} (-1)^r x^{12 - 3r}$$

Set the exponent of $x$ equal to $0$: $12 - 3r = 0 \implies r = 4$. Substitute $r = 4$:

$$\text{Constant Term} = \binom{6}{4} \cdot 2^{6-4} \cdot (-1)^4 = 15 \cdot 4 \cdot 1 = 60$$

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Worked Example 2: Geometric Combinations & Conditional Probability

Problem:

1. In a plane, $9$ points are given such that $4$ of them lie on a straight line $d$, and the remaining $5$ points lie on a circle (with no other collinear triples). How many triangles can be formed using these $9$ points as vertices?
2. Two fair dice are rolled and it is known that the sum of the top faces is $8$. What is the probability that at least one of the dice shows a $3$?

Step-by-Step Solution:

1. Choosing any $3$ points out of $9$ gives $\binom{9}{3}$ triples, while choosing $3$ collinear points from line $d$ gives $\binom{4}{3}$ degenerate triples:

$$\text{Triangles} = \binom{9}{3} - \binom{4}{3} = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} - 4 = 84 - 4 = 80$$

1. Let $B$ be the given condition that the sum of the two dice is $8$: $B = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}$, so $s(B) = 5$. Let $A$ be the event that at least one die is a $3$. Then $A \cap B = \{(3,5), (5,3)\}$, so $s(A \cap B) = 2$. Thus:

$$P(A \mid B) = \frac{s(A \cap B)}{s(B)} = \frac{2}{5}$$

Conditional probability requires $P(B)>0$. The circular counts assume distinct objects and that rotations are equivalent; labelled fixed seats change the count. A flip-symmetric keyring additionally identifies reversed cyclic orders and uses the displayed halving formula for $n\ge3$.

Review the Method

For probability, P(A U B) = P(A) + P(B) - P(A n B), independent events satisfy P(A n B) = P(A)P(B), conditional probability is P(A | B) = P(A n B) / P(B), and 'at least one' problems are fastest via 1 - P(none).