- #1
SithsNGiggles
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Homework Statement
I just need to know if it makes sense; I was told that I can't have anyone make any improvements on what I've written myself.
Question: If [itex]X = \{ x_1 , \ldots, x_n \}[/itex] and [itex]Y = \{ y_1 , \ldots, y_m \}[/itex], how many functions from [itex]X[/itex] to [itex]Y[/itex] exist?
My answer: [itex]m^n[/itex] functions
The Attempt at a Solution
For any element [itex]x \in X[/itex], there exists a unique [itex]y \in Y[/itex] for which [itex]F(x) = y[/itex].
Every [itex]n[/itex] element in [itex[X[/itex] will be paired with anyone of the [itex]m[/itex] elements in [itex]Y[/itex].
i.e. there exist [itex]m[/itex] possible [itex]F(x_1)[/itex] in [itex]Y[/itex] that can be paired with [itex]x_1[/itex].
[itex]x_2[/itex] can be paired with [itex]m[/itex] possible [itex]F(x_2)[/itex]
[itex]\vdots[/itex]
[itex]x_n[/itex] can be paired with [itex]m[/itex] possible [itex]F(x_n)[/itex].
Because the domain [itex]D_F = X[/itex], every function generated through F will contain [itex]n[/itex] coordinate pairs. Furthermore, since there are [itex]m[/itex] possible values [itex]F(x) = y[/itex] for each element [itex]x[/itex], there are [itex]n[/itex] factors of [itex]m[/itex], or [itex]m^n[/itex], possible functions.
Thanks for any commentary (but not actual help!) you can provide.