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(b) For a set X, let (*) denote the set of subsets of X of cardinality k. Given n>1 we define three sets of cardinality n: X₂ = {1,...,n}, X {1,..., n'} and X = {1",...,n"}. For n>3 construct a bijection X" • (X - ²) U (X^ 3) u (X² 3). U 3 f: : (₂X2) → and prove that it is a bijection.

(b) For a set X, let (*) denote the set of subsets of X of cardinality k. Given n>1 we define three sets of cardinality n: X₂ = {1,...,n}, X {1,..., n'} and X = {1",...,n"}. For n>3 construct a bijection X" • (X - ²) U (X^ 3) u (X² 3). U 3 f: : (₂X2) → and prove that it is a bijection.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.6: Inequalities
Problem 78E
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(b) For a set X, let (X) denote the set of subsets of X of cardinality k. Given n ≥ 1 we define three sets of cardinality n: X₂ = {1,...,n}, X {1,..., n'} and X = {1",...,n"}. For n>3 construct a bijection 2 f: : (₂x^2) - • (X - 3) U (X^ 3) U (XH 3), n- and prove that it is a bijection.
Transcribed Image Text:(b) For a set X, let (X) denote the set of subsets of X of cardinality k. Given n ≥ 1 we define three sets of cardinality n: X₂ = {1,...,n}, X {1,..., n'} and X = {1",...,n"}. For n>3 construct a bijection 2 f: : (₂x^2) - • (X - 3) U (X^ 3) U (XH 3), n- and prove that it is a bijection.
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