Erratum

This article refers to:

Derivation of some formulae in combinatrics by heuristic methods

Kobayashi, Y. (2015). Derivation of some formulae in combinatrics by heuristic methods. International Journal of Mathematical Education in Science and Technology, 46(3), pp.469–476. http://dx.doi.org/10.1080/0020739X.2014.982735.

In the print version, an equation was mistakenly presented on p.475 instead of p.474. The correct position of this equation in relation to the rest of the text is presented below.

Changing our point of view, let us consider the subsets of {1, 2, … , n}. For simplicity, the case of n = 4 is shown below.

Table

		${\underbrace{1\le 1\le 1\le 1}}_{1\mathrm{way}}$, ${\underbrace{2\le 2\le 2\le 2}}_{1\mathrm{way}}$,
1 element	${\underbrace{\left\{1\right\},\left\{2\right\},\left\{3\right\},\left\{4\right\}}}_{{}_4{C}_1\mathrm{ways}}$	${\underbrace{3\le 3\le 3\le 3}}_{1\mathrm{way}}$, ${\underbrace{4\le 4\le 4\le 4}}_{1\mathrm{way}}$	(4C1×1)   ways
		$1\le \le \le 2$, … ,
2 elements	${\underbrace{\{1,2\},...,\{3,4\}}}_{{}_4{C}_2\mathrm{ways}}$	$3\le \le \le 4$	(4C2×3) ways
		$1-1$ $3-3$
		$1-2$ $3-4$
		${\underbrace{2-2}}_{3\mathrm{ways}}$ ${\underbrace{4-4}}_{3\mathrm{ways}}$
		$1\le \le \le 3$, … ,
3 elements	${\underbrace{\{1,2,3\},...,\{2,3,4\}}}_{{}_4{C}_3\mathrm{ways}}$	$2\le \le \le 4$	(4C3×3) ways
		$1-2$ $2-3$
		$2-2$ $3-3$
		${\underbrace{2-3}}_{3\mathrm{ways}}$ ${\underbrace{3-4}}_{3\mathrm{ways}}$
4 elements	${\underbrace{\{1,2,3,4\}}}_{{}_4{C}_4\mathrm{way}}$	${\underbrace{1\le 2\le 3\le 4}}_{1\mathrm{way}}$	(4C4×1) way.

Thus, the number of mappings satisfying f(1) ≤ f(2) ≤ f(3) ≤ f(4) is ₄C₁ × 1 + ₄C₂ × 3 + ₄C₃ × 3 + ₄C₄ × 1.

Taylor & Francis apologises for this error.