Let \(A\) and \(B\) be the image of \(f\) and \(f \circ f\)
respectively.
If \(n(A) = 6\), then \(f\) is bijective and \(f \circ f\) is also
bijective, so \(n(B) = 6\).
If \(n(A) \leq 4\) then \(n(B) \leq 4\) since \(B \subseteq A\).
Therefore it should be that \(n(A) = 5\) and \(B = A\).
-
There are \(\fbox{\(\;(\alpha)\;\)}\) ways to select a subset
\(A\) of \(X\) where \(n(A) = 5\).
-
For the set \(A\) selected in (i),
let \(k\) be the element of \(X\) that is not in \(A\). Since
\(n(A) = 5\), there are \(\fbox{\(\;(\beta)\;\)}\) ways to
select \(f(k)\) from \(A\).
-
For \(A=\{a_1,a_2,a_3,a_4,a_5\}\) selected in
(i) and \(f(k)\) selected in
(ii),
since \(f(k) \in A\) and \(A = B\), it should be that
\(A=\{f(a_1),f(a_2),f(a_3),f(a_4),f(a_5)\}\).
The number of cases where this holds is equal to the number of
bijections from \(A\) to \(A\), which is
\(\fbox{\(\;(\gamma)\;\)}\).
By
(i), (ii) and
(iii), the number of functions \(f\)
to compute is
\(\fbox{\(\;(\alpha)\;\)}\times\fbox{\(\;(\beta)\;\)}\times\fbox{\(\;(\gamma)\;\)}\).
shape inside the triangle \(\mathrm{OA_1B_1}\) and the sector
\(\mathrm{OA_1B_2}\) except the overlapping parts. Starting from
figure \(R_1\), let \(\mathrm{A_2}\) be the point where the line
\(\mathrm{OA_1}\) meets a line passing through point \(\mathrm{B_2}\)
parallel to line \(\mathrm{A_1B_1}\). Let \(\mathrm{B_3}\) be the
point where the line segment \(\mathrm{OB_2}\) meets a circle with
center \(\mathrm{O}\) and radius \(\overline{\mathrm{OA_2}}\). Obtain
figure \(R_2\) by coloring the
