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Question #41064

Difficulty: 8/10 | 🟢 Free

The Question

Let $A = \{1, 2, 3, 4, 5, 6\}$ and $B = \{1, 2, 3, \dots, 14\}$ . The number of injective functions $f: A \to B$ such that $f(1) + f(2) = 15$ , and for all two-element subsets $\{i, j\} \subseteq A$ other than $\{1, 2\}$ , $f(i) + f(j) \neq 15$ , is:

Options

A.
80640
B.
166320
C.
40320
D.
5040

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Combinatorial Counting (P&C Integration)|Injective Mapping (One-One)|Domain Restrictions|Domain Constraint Intersections

Why This Question Is Tricky

Common Mistakes

The core trap in this problem is the 'phantom constraint' intersection.

Students are given a global restriction ( $f(i) + f(j) \neq 15$ ) with a single exception ( $f(1) + f(2) = 15$ ).

A common mistake is attempting complex inclusion-exclusion to ensure $f(3), f(4), f(5), f(6)$ do not sum to 15 with $f(1)$ or $f(2)$ .

However, the injective property inherently guarantees this: if $f(3) + f(1) = 15$ , then $f(3)$ must equal $f(2)$ , which violates injectivity.

By recognizing that the cross-terms are automatically handled by the injective constraint, the problem elegantly reduces to selecting 4 elements from the remaining 6 disjoint pairs in the codomain such that no two elements share the same pair, testing both state-space reduction and combinatorial pair-picking.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Combinatorial Counting (P&C Integration)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free