Fakultät/10/Zerlegung/Aufgabe/Lösung
<
Fakultät/10/Zerlegung/Aufgabe
Es ist
10
!
=
10
⋅
9
⋅
8
⋅
(
7
!
)
=
2
⋅
5
⋅
3
⋅
3
⋅
4
⋅
2
⋅
(
7
!
)
=
(
2
⋅
3
)
⋅
5
⋅
4
⋅
3
⋅
2
⋅
(
7
!
)
=
(
6
!
)
⋅
(
7
!
)
.
{\displaystyle {}{\begin{aligned}10!&=10\cdot 9\cdot 8\cdot (7!)\\&=2\cdot 5\cdot 3\cdot 3\cdot 4\cdot 2\cdot (7!)\\&=(2\cdot 3)\cdot 5\cdot 4\cdot 3\cdot 2\cdot (7!)\\&=(6!)\cdot (7!).\end{aligned}}}
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