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Tardigrade
Question
Mathematics
If x is a positive integer, then | x! (x+1)! (x+2)! [0.3em] (x+1)! (x+2)! (x+3)! [0.3em] (x+2)! (x+3)! &(x+4)! | is equal to
Q. If x is a positive integer, then
∣
∣
x
!
(
x
+
1
)!
(
x
+
2
)!
(
x
+
1
)!
(
x
+
2
)!
(
x
+
3
)!
(
x
+
2
)!
(
x
+
3
)!
(
x
+
4
)!
∣
∣
is equal to
2163
200
Determinants
Report Error
A
2
x
!
(
x
+
1
)!
8%
B
2
x
!
(
x
+
1
)!
(
x
+
2
)!
46%
C
2
x
!
(
x
+
3
)!
15%
D
2
(
x
+
1
)!
(
x
+
2
)!
(
x
+
3
)!
30%
Solution:
∣
∣
x
!
(
x
+
1
)!
(
x
+
2
)!
(
x
+
1
)!
(
x
+
2
)!
(
x
+
3
)!
(
x
+
2
)!
(
x
+
3
)!
(
x
+
4
)!
∣
∣
=
x
!
(
x
+
1
)!
(
x
+
2
)!
∣
∣
1
x
+
1
(
x
+
2
)
(
x
+
1
)
1
x
+
2
(
x
+
2
)
(
x
+
3
)
1
(
x
+
3
)
(
x
+
3
)
(
x
+
4
)
∣
∣
=
2
x
!
(
x
+
1
)!
(
x
+
2
)!