Study guide
5.2 Combinatorics: Permutations, Combinations & Factorials
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1 Chapter 1: Exam Structure, Scoring Mechanics & Test Strategy
1.1 Test Architecture, Timing & the Sequential Test Interface 1.2 Scoring System, Penalty Mathematics & Strategic Answering
2 Chapter 2: Mathematics — Algebra, Equations & Inequalities
2.1 Algebraic Expressions, Polynomials, Powers & Radicals 2.2 Linear & Quadratic Equations and Systems 2.3 Linear & Quadratic Inequalities and Sign Charts 2.4 Equations and Inequalities with Absolute Values
3 Chapter 3: Mathematics — Functions, Exponents & Logarithms
3.1 Function Concepts, Domain, Range & Inverses 3.2 Composition of Functions & Transformations 3.3 Exponential Functions, Properties & Equations 3.4 Logarithmic Functions, Laws & Equations
4 Chapter 4: Mathematics — Geometry, Analytic Geometry & Trigonometry
4.1 Plane Geometry: Polygons, Circles & Area Theorems 4.2 Analytic Geometry: Lines, Slopes & Distances in the Cartesian Plane 4.3 Conic Sections: Circles & Parabolas in the Coordinate Plane 4.4 Trigonometric Functions, Identities & Right-Triangle Applications
5 Chapter 5: Mathematics — Combinatorics, Probability, Statistics & Set Theory
5.1 Set Theory, Operations & Venn Diagrams 5.2 Combinatorics: Permutations, Combinations & Factorials 5.3 Probability Theory: Independent Events, Conditional Probability & Bayes' Basics 5.4 Descriptive Statistics: Central Tendency, Dispersion & Frequency Distributions
6 Chapter 6: Reading Comprehension — Business, Economic & Academic Analysis
6.1 Main Idea, Text Structure & Passage Mapping 6.2 Explicit Information Retrieval & Detail Verification 6.3 Inference, Implicit Meaning & Authorial Purpose 6.4 Tone, Rhetorical Devices & Vocabulary in Academic Context
7 Chapter 7: Numerical Reasoning — Charts, Tables & Data Interpretation
7.1 Data Tables & Multi-Series Matrices 7.2 Graphical Interpretation: Bar, Line, Pie & Composite Charts 7.3 Percentage Calculations, Changes & Financial Metrics 7.4 Ratios, Proportions, Rates & Weighted Averages
8 Chapter 8: Logic & Critical Thinking — Formal & Deductive Reasoning
8.1 Propositional Logic, Truth Tables & Logical Connectives 8.2 Conditional Reasoning, Contrapositives & Modus Ponens/Tollens 8.3 Syllogisms, Quantifiers & Venn Logic
9 Chapter 9: Critical Thinking — Argument Analysis, Assumptions & the Official Item Types
9.1 Argument Components, Premises & Unstated Assumptions 9.2 Strengthening, Weakening & Flaw Identification 9.3 Constraint Deduction Drills: Ordering, Arrangement & Assignment 9.4 The Two Official Critical-Thinking Item Types: Data Propositions & Deducibility
5.2 Combinatorics: Permutations, Combinations & Factorials
Key Takeaways Explain This Topic AI
The Fundamental Counting Principle establishes that sequential independent choices with n₁, n₂, ..., nₖ options yield n₁ · n₂ · ... · nₖ total outcomes.
Permutations P(n, k) = n! / (n - k)! arrange distinct objects where order matters and combinations C(n, k) = n! / [k!(n - k)!] select unordered subsets, while combinations with repetition C(n + k - 1, k) — stars and bars — complete the four-cell order-versus-repetition table.
Permutations of multisets with repeated items are governed by the multinomial formula n! / (n₁! · n₂! · ... · nₖ!), essential for letter anagrams.
Pascal's Identity C(n, k) + C(n, k - 1) = C(n + 1, k) and symmetry C(n, k) = C(n, n - k) enable rapid mental reduction of large binomial coefficients.
Circular arrangements eliminate rotational duplicates, yielding (n - 1)! distinct orderings (or (n - 1)! / 2 when reflective symmetry applies, as in keyrings).
Last updated: September 2026
5.2 Combinatorics: Permutations, Combinations & Factorials
Combinatorics is the branch of discrete mathematics dedicated to counting, arranging, and selecting objects according to specified rules. On the Bocconi Admission Test, combinatorial questions appear both as direct counting problems and as the foundational sample-space calculations required for probability items. Under the pressure of a 90-second pacing constraint, candidates must instantly classify whether a problem requires permutations (where order matters) or combinations (where order is irrelevant), apply algebraic simplifications to factorials, and enforce structural constraints such as grouping or complementary counting.
1. Fundamental Principles of Counting
All combinatorial mathematics rests upon two fundamental logical rules:
The Rule of Product (Multiplication Principle)
If a multi-stage procedure consists of k k k sequential, independent decisions such that the first decision can be made in n 1 n_1 n 1 ways, the second in n 2 n_2 n 2 ways, ..., and the k k k -th in n k n_k n k ways, then the total number of distinct outcomes is: N = n 1 ⋅ n 2 ⋅ n 3 ⋯ n k N = n_1 \cdot n_2 \cdot n_3 \cdots n_k N = n 1 ⋅ n 2 ⋅ n 3 ⋯ n k
Example: An online Bocconi exam PIN consists of 2 uppercase letters followed by 3 digits. If repetition is allowed, the total number of unique PINs is 26 × 26 × 10 × 10 × 10 = 676,000 26 \times 26 \times 10 \times 10 \times 10 = 676{,}000 26 × 26 × 10 × 10 × 10 = 676 , 000 .
The Rule of Sum (Addition Principle)
If a single choice must be made from either set A A A or set B B B , and the two sets are mutually exclusive ( A ∩ B = ∅ A \cap B = \emptyset A ∩ B = ∅ ), with ∣ A ∣ = m |A| = m ∣ A ∣ = m and ∣ B ∣ = n |B| = n ∣ B ∣ = n , then the number of possible choices is: N = ∣ A ∣ + ∣ B ∣ = m + n N = |A| + |B| = m + n N = ∣ A ∣ + ∣ B ∣ = m + n
If the sets overlap, we must apply the Principle of Inclusion-Exclusion: N = ∣ A ∣ + ∣ B ∣ − ∣ A ∩ B ∣ N = |A| + |B| - |A \cap B| N = ∣ A ∣ + ∣ B ∣ − ∣ A ∩ B ∣ .
2. Factorials and Operational Properties
For any positive integer n ∈ N n \in \mathbb{N} n ∈ N , the factorial n ! n! n ! represents the product of all positive integers less than or equal to n n n : n ! = n ⋅ ( n − 1 ) ⋅ ( n − 2 ) ⋯ 3 ⋅ 2 ⋅ 1 n! = n \cdot (n - 1) \cdot (n - 2) \cdots 3 \cdot 2 \cdot 1 n ! = n ⋅ ( n − 1 ) ⋅ ( n − 2 ) ⋯ 3 ⋅ 2 ⋅ 1
Conventions and Algebraic Shortcuts
Zero Factorial: By mathematical convention (the empty product), 0 ! = 1 0! = 1 0 ! = 1 . This ensures formulas such as C ( n , n ) = n ! n ! 0 ! = 1 C(n, n) = \frac{n!}{n! 0!} = 1 C ( n , n ) = n ! 0 ! n ! = 1 hold consistently.
Factorial Ratios: Always cancel common terms before multiplying: n ! ( n − 1 ) ! = n , n ! ( n − 2 ) ! = n ( n − 1 ) = n 2 − n , ( n + 1 ) ! ( n − 1 ) ! = ( n + 1 ) n = n 2 + n \frac{n!}{(n - 1)!} = n, \qquad \frac{n!}{(n - 2)!} = n(n - 1) = n^2 - n, \qquad \frac{(n + 1)!}{(n - 1)!} = (n + 1)n = n^2 + n ( n − 1 )! n ! = n , ( n − 2 )! n ! = n ( n − 1 ) = n 2 − n , ( n − 1 )! ( n + 1 )! = ( n + 1 ) n = n 2 + n Mental calculation example: 12 ! 10 ! ⋅ 2 ! = 12 ⋅ 11 2 ⋅ 1 = 6 ⋅ 11 = 66 \frac{12!}{10! \cdot 2!} = \frac{12 \cdot 11}{2 \cdot 1} = 6 \cdot 11 = 66 10 ! ⋅ 2 ! 12 ! = 2 ⋅ 1 12 ⋅ 11 = 6 ⋅ 11 = 66 .
3. Permutations: Order Matters
A permutation is an ordered arrangement of objects. The sequence in which elements appear defines a distinct outcome.
Linear Permutations Without Repetition
Arranging n n n distinct objects into n n n positions: P ( n ) = n ! P(n) = n! P ( n ) = n !
Arranging k k k objects selected from n n n distinct items ( 0 ≤ k ≤ n 0 \le k \le n 0 ≤ k ≤ n ): P ( n , k ) = n P k = n ! ( n − k ) ! = n ( n − 1 ) ( n − 2 ) ⋯ ( n − k + 1 ) P(n, k) = {}_n P_k = \frac{n!}{(n - k)!} = n(n - 1)(n - 2)\cdots(n - k + 1) P ( n , k ) = n P k = ( n − k )! n ! = n ( n − 1 ) ( n − 2 ) ⋯ ( n − k + 1 ) Speed note: To evaluate P ( 8 , 3 ) P(8, 3) P ( 8 , 3 ) , multiply 8 downwards for 3 terms: 8 × 7 × 6 = 336 8 \times 7 \times 6 = 336 8 × 7 × 6 = 336 .
Linear Permutations With Repetition (Multisets / Anagrams)
When arranging n n n total objects where some objects are indistinguishable, dividing by the factorials of the repeated counts eliminates duplicate orderings. If there are n 1 n_1 n 1 items of type 1, n 2 n_2 n 2 of type 2, ..., n r n_r n r of type r r r (with n 1 + n 2 + ⋯ + n r = n n_1 + n_2 + \dots + n_r = n n 1 + n 2 + ⋯ + n r = n ): P ( n ; n 1 , n 2 , … , n r ) = n ! n 1 ! ⋅ n 2 ! ⋯ n r ! P(n; n_1, n_2, \dots, n_r) = \frac{n!}{n_1! \cdot n_2! \cdots n_r!} P ( n ; n 1 , n 2 , … , n r ) = n 1 ! ⋅ n 2 ! ⋯ n r ! n !
Classic Application: The number of distinct anagrams of the word STATISTICS (10 letters: 3 S, 3 T, 2 I, 1 A, 1 C) is: 10 ! 3 ! ⋅ 3 ! ⋅ 2 ! ⋅ 1 ! ⋅ 1 ! = 3,628,800 6 ⋅ 6 ⋅ 2 = 50,400 \frac{10!}{3! \cdot 3! \cdot 2! \cdot 1! \cdot 1!} = \frac{3{,}628{,}800}{6 \cdot 6 \cdot 2} = 50{,}400 3 ! ⋅ 3 ! ⋅ 2 ! ⋅ 1 ! ⋅ 1 ! 10 ! = 6 ⋅ 6 ⋅ 2 3 , 628 , 800 = 50 , 400
Permutations With Replacement
Selecting k k k objects from n n n distinct types where an object can be reused indefinitely yields: N = n k N = n^k N = n k
Circular Permutations
Arranging n n n distinct items around a circle eliminates absolute positional labels. Because rotating all items by one seat preserves identical relative neighbors, dividing by n n n rotational shifts gives: P circ ( n ) = n ! n = ( n − 1 ) ! P_{\text{circ}}(n) = \frac{n!}{n} = (n - 1)! P circ ( n ) = n n ! = ( n − 1 )!
Reflective Symmetry (Necklaces and Keychains): If flipping the circle over produces an identical arrangement (no distinct clockwise vs. counterclockwise orientation), divide by an additional factor of 2: P necklace ( n ) = ( n − 1 ) ! 2 P_{\text{necklace}}(n) = \frac{(n - 1)!}{2} P necklace ( n ) = 2 ( n − 1 )!
4. Combinations: Order Does Not Matter
A combination is an unordered selection of k k k elements from a set of n n n distinct objects. The internal sequence of chosen elements is irrelevant.
Formula for Combinations C ( n , k ) = ( n k ) = P ( n , k ) k ! = n ! k ! ( n − k ) ! C(n, k) = \binom{n}{k} = \frac{P(n, k)}{k!} = \frac{n!}{k!(n - k)!} C ( n , k ) = ( k n ) = k ! P ( n , k ) = k ! ( n − k )! n !
Algebraic Properties of Binomial Coefficients
Symmetry Identity: ( n k ) = ( n n − k ) \binom{n}{k} = \binom{n}{n - k} ( k n ) = ( n − k n ) Fast computation: Instead of computing ( 12 10 ) \binom{12}{10} ( 10 12 ) , compute ( 12 2 ) = 12 ⋅ 11 2 ⋅ 1 = 66 \binom{12}{2} = \frac{12 \cdot 11}{2 \cdot 1} = 66 ( 2 12 ) = 2 ⋅ 1 12 ⋅ 11 = 66 .
Boundary Values: ( n 0 ) = 1 , ( n n ) = 1 , ( n 1 ) = n , ( n n − 1 ) = n \binom{n}{0} = 1, \quad \binom{n}{n} = 1, \quad \binom{n}{1} = n, \quad \binom{n}{n - 1} = n ( 0 n ) = 1 , ( n n ) = 1 , ( 1 n ) = n , ( n − 1 n ) = n
Pascal's Identity: ( n k ) + ( n k − 1 ) = ( n + 1 k ) \binom{n}{k} + \binom{n}{k - 1} = \binom{n + 1}{k} ( k n ) + ( k − 1 n ) = ( k n + 1 )
Sum of Row in Pascal's Triangle: ∑ k = 0 n ( n k ) = ( n 0 ) + ( n 1 ) + ⋯ + ( n n ) = 2 n \sum_{k=0}^n \binom{n}{k} = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n k = 0 ∑ n ( k n ) = ( 0 n ) + ( 1 n ) + ⋯ + ( n n ) = 2 n Significance: This sum equals the total cardinality of the power set P ( S ) \mathcal{P}(S) P ( S ) of an n n n -element set.
Combinations With Repetition (Multisets)
Bocconi's syllabus asks for "the main formulas of combinatorics (permutations, combinations with and without repetitions )", so the repetition case is examinable in its own right. The situation is: choose k k k items from n n n distinct types , where each type may be chosen any number of times and the order of selection is irrelevant . C R ( n , k ) = ( n + k − 1 k ) = ( n + k − 1 ) ! k ! ( n − 1 ) ! C^{R}(n, k) = \binom{n + k - 1}{k} = \frac{(n + k - 1)!}{k! \, (n - 1)!} C R ( n , k ) = ( k n + k − 1 ) = k ! ( n − 1 )! ( n + k − 1 )!
Why the formula looks like that (stars and bars). Represent a selection as k k k identical stars distributed among n n n type-buckets, with n − 1 n - 1 n − 1 bars separating the buckets. Any arrangement of the k k k stars and n − 1 n - 1 n − 1 bars encodes exactly one selection, so the count is the number of ways of choosing which k k k of the n + k − 1 n + k - 1 n + k − 1 positions hold stars.
Worked Example. A café sells 4 kinds of pastry. A customer buys 6 pastries, and may repeat kinds freely. How many different orders are possible? C R ( 4 , 6 ) = ( 4 + 6 − 1 6 ) = ( 9 6 ) = ( 9 3 ) = 9 × 8 × 7 3 × 2 × 1 = 84 C^{R}(4, 6) = \binom{4 + 6 - 1}{6} = \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 C R ( 4 , 6 ) = ( 6 4 + 6 − 1 ) = ( 6 9 ) = ( 3 9 ) = 3 × 2 × 1 9 × 8 × 7 = 84
The Four-Cell Counting Table
Classifying a counting problem correctly is worth more than knowing any single formula. Two binary questions settle it: does order matter, and is repetition allowed?
Repetition NOT allowed Repetition allowed
Order matters (permutations) P ( n , k ) = n ! ( n − k ) ! P(n, k) = \dfrac{n!}{(n-k)!} P ( n , k ) = ( n − k )! n ! n k n^{k} n k
Order does not matter (combinations) ( n k ) \binom{n}{k} ( k n ) ( n + k − 1 k ) \dbinom{n + k - 1}{k} ( k n + k − 1 )
Bocconi Trap: the phrase "how many different selections" almost always means order does not matter, while "how many different arrangements", "codes" or "sequences" means it does. Read for the noun, not the verb.
5. The Binomial Theorem
For any real numbers a , b a, b a , b and non-negative integer n n n : ( a + b ) n = ∑ k = 0 n ( n k ) a n − k b k = ( n 0 ) a n + ( n 1 ) a n − 1 b + ⋯ + ( n n ) b n (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n - k} b^k = \binom{n}{0} a^n + \binom{n}{1} a^{n - 1} b + \dots + \binom{n}{n} b^n ( a + b ) n = k = 0 ∑ n ( k n ) a n − k b k = ( 0 n ) a n + ( 1 n ) a n − 1 b + ⋯ + ( n n ) b n
General ( k + 1 ) (k + 1) ( k + 1 ) -th Term
The general term in the binomial expansion is given by: T k + 1 = ( n k ) a n − k b k T_{k+1} = \binom{n}{k} a^{n - k} b^k T k + 1 = ( k n ) a n − k b k
Determining Specific Terms (e.g., Constant Term)
To find the term independent of x x x (the constant term) in an expansion such as ( x a + c x b ) n \left(x^a + \frac{c}{x^b}\right)^n ( x a + x b c ) n :
Express the general term with a a a and b b b substituted.
Collect the powers of x x x : x ( n − k ) a − k b x^{(n - k)a - kb} x ( n − k ) a − kb .
Set the net exponent to zero: ( n − k ) a − k b = 0 (n - k)a - kb = 0 ( n − k ) a − kb = 0 , and solve for integer k k k .
Substitute k k k back into T k + 1 T_{k+1} T k + 1 to evaluate the numeric coefficient.
6. Constrained Counting & Committee Problems
Bocconi exam questions frequently superimpose operational constraints on selections:
The 'At Least One' Principle (Complementary Counting)
When asked to find the number of arrangements or selections containing 'at least one' of a specific type, avoid summing multiple disjoint cases. Instead, subtract the complement from the total: N ( at least one ) = N ( total ) − N ( none ) N(\text{at least one}) = N(\text{total}) - N(\text{none}) N ( at least one ) = N ( total ) − N ( none )
The 'Block Method' (Together / Adjacent)
When specific objects must always remain adjacent, treat them as a single fused 'super-item':
Count arrangements of the remaining objects plus the single super-item.
Multiply by the internal permutations of the objects within the super-item.
The 'Gap Method' (Separated / Never Adjacent)
When specific objects must never be adjacent:
Arrange all unconstrained items first, creating spaces (gaps) before, between, and after them.
Choose gaps from the available spaces to place the restricted items using combinations C ( gaps , k ) C(\text{gaps}, k) C ( gaps , k ) .
7. Common Bocconi Traps & Fast Shortcuts
The Order Fallacy: Confusing permutations with combinations. Always ask: Does changing the order of the selected items create a different valid outcome? If selecting a committee of co-equal roles ⟹ \implies ⟹ Combination. If assigning specific roles (e.g., President, Vice-President, Treasurer) ⟹ \implies ⟹ Permutation.
The Partial Selection Double-Count Trap: When selecting 'at least 2 women' for a committee of 4 from a pool of 5 women and 5 men, NEVER select 2 women first ( 5 2 ) \binom{5}{2} ( 2 5 ) and then choose the remaining 2 members from the remaining 8 people ( 8 2 ) \binom{8}{2} ( 2 8 ) . This massively overcounts because a committee with women { W 1 , W 2 , W 3 , M 1 } \{W_1, W_2, W_3, M_1\} { W 1 , W 2 , W 3 , M 1 } is counted multiple times depending on which women were chosen in the first step. Always partition into disjoint cases or use complementary counting.
Cancellation Before Multiplication: Never multiply factorials out into large numbers. In ( 10 4 ) = 10 × 9 × 8 × 7 4 × 3 × 2 × 1 \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} ( 4 10 ) = 4 × 3 × 2 × 1 10 × 9 × 8 × 7 , immediately cancel 4 × 2 4 \times 2 4 × 2 with 8 8 8 , and divide 9 9 9 by 3 3 3 to get 3 3 3 , leaving 10 × 3 × 7 = 210 10 \times 3 \times 7 = 210 10 × 3 × 7 = 210 .
- 8. Worked Step-by-Step Examples
- Example 1: Multiset Permutations with Adjacency Constraints
How many 7-letter anagrams can be formed from the word BOCCONI such that the two letters 'C' are always adjacent?
Character Frequency: Letters are { B : 1 , O : 2 , C : 2 , N : 1 , I : 1 } \{B:1, O:2, C:2, N:1, I:1\} { B : 1 , O : 2 , C : 2 , N : 1 , I : 1 } . Total letters = 7.
Apply the Block Method: Fuse the two 'C's into a single super-item [ C C ] [CC] [ CC ] .
Count Items: The items to arrange are now { B , O , O , N , I , [ C C ] } \{B, O, O, N, I, [CC]\} { B , O , O , N , I , [ CC ]} , which equals 6 items.
Multiset Permutation of 6 Items: Among these 6 items, 'O' appears twice. Thus: N = 6 ! 2 ! = 720 2 = 360 N = \frac{6!}{2!} = \frac{720}{2} = 360 N = 2 ! 6 ! = 2 720 = 360
Internal Permutations: Because both letters in [ C C ] [CC] [ CC ] are identical, there is only 2 ! 2 ! = 1 \frac{2!}{2!} = 1 2 ! 2 ! = 1 internal arrangement.
Hence, exactly 360 valid anagrams can be formed.
Example 2: Committee Selection with Subgroup Constraints
A corporate advisory board consists of 6 economists and 4 lawyers. A committee of 4 members must be chosen. How many ways can the committee be formed if it must contain at least one lawyer?
Total Selections Without Restriction: Total candidates = 6 + 4 = 10 6 + 4 = 10 6 + 4 = 10 . N total = ( 10 4 ) = 10 × 9 × 8 × 7 4 × 3 × 2 × 1 = 210 N_{\text{total}} = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 N total = ( 4 10 ) = 4 × 3 × 2 × 1 10 × 9 × 8 × 7 = 210
Selections with No Lawyers (All Economists): Choose all 4 members from the 6 economists: N none = ( 6 4 ) = ( 6 2 ) = 6 × 5 2 × 1 = 15 N_{\text{none}} = \binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 N none = ( 4 6 ) = ( 2 6 ) = 2 × 1 6 × 5 = 15
Apply Complementary Subtraction: N = N total − N none = 210 − 15 = 195 N = N_{\text{total}} - N_{\text{none}} = 210 - 15 = 195 N = N total − N none = 210 − 15 = 195
Example 3: Constant Term in a Binomial Expansion
Find the term independent of x x x in the algebraic expansion of ( x 2 − 2 x ) 6 \left(x^2 - \frac{2}{x}\right)^6 ( x 2 − x 2 ) 6 .
Write General Term: T k + 1 = ( 6 k ) ( x 2 ) 6 − k ( − 2 x ) k = ( 6 k ) x 12 − 2 k ( − 2 ) k x − k = ( 6 k ) ( − 2 ) k x 12 − 3 k T_{k+1} = \binom{6}{k} (x^2)^{6 - k} \left(-\frac{2}{x}\right)^k = \binom{6}{k} x^{12 - 2k} (-2)^k x^{-k} = \binom{6}{k} (-2)^k x^{12 - 3k} T k + 1 = ( k 6 ) ( x 2 ) 6 − k ( − x 2 ) k = ( k 6 ) x 12 − 2 k ( − 2 ) k x − k = ( k 6 ) ( − 2 ) k x 12 − 3 k
Enforce Constant Term Exponent: 12 − 3 k = 0 ⟹ 3 k = 12 ⟹ k = 4 12 - 3k = 0 \implies 3k = 12 \implies k = 4 12 − 3 k = 0 ⟹ 3 k = 12 ⟹ k = 4
Evaluate Coefficient: T 5 = ( 6 4 ) ( − 2 ) 4 = ( 6 2 ) ⋅ 16 = 15 ⋅ 16 = 240 T_5 = \binom{6}{4} (-2)^4 = \binom{6}{2} \cdot 16 = 15 \cdot 16 = 240 T 5 = ( 4 6 ) ( − 2 ) 4 = ( 2 6 ) ⋅ 16 = 15 ⋅ 16 = 240 The constant term is 240.