Journal of Science and Arts Year 13, No. 3(24), pp. 217-230, 2013
ISSN: 1844 – 9581 Mathematics Section
ORIGINAL PAPER
SOME NEW IDENTITIES IN COMBINATORIC
DAM VAN NHI1
_________________________________________________
, TRAN TRUNG TINH1
Manuscript received15.07.2013; Accepted paper: 28.07.2013;
Published online: 15,09.2013.
Abstract. In this paper we introduce some new identities in combinatoric.
Keywords: Equation, Identity, Combinatoric.
2010 Mathematics Subject Classification: 26D05, 26D15, 51M16.
1. INTRODUCTION
Proposition 1.1. Denote
( ) ( )
1
n
i
i
xx
ϕα
=
= +
∏
. Then, there are the following identities:
(i)
( ) ( ) ( )
0
1 1!
n
in
i
n
in
i
ϕ
=

−=−


∑.
(ii)
( ) ( )
0
1 1!
nin
n
i
nin
i
=

−=−


∑
.
(iii)
( ) ( )
0
1
1 1.
=
+

−=−


∑
nin
i
nn
ii
(iv)
( ) ( ) ( ) ( )
1
14! 10
21
i
nn
i
ni
in
ic
ϕ
ϕ
=

−
 = +−
−
∑
.
(v)
( ) ( )
()
12
1
1 2!
2 1 2!
in
n
n
i
nn
i
i
in
−
=
− =

−
∑
.
Proposition 1.2. There is the following identity:
( )
( )
( )
2 22
1
2
0
2
1
2 2!
1
n
nk
n
s
k
n
k sknk n
k
−
=
=

−+

−

=
+
∏
∑
.
1 Ha Noi National University of Education, 136 Xuan Thuy Road, Cau Giay, Hanoi, Vietnam.
E-mail: damvannhi@yahoo.com.
Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh
www.josa.ro Mathematics Section
Proposition 1.3. For all integer
1n≥
, there is the following identity:
( )
( )
( )
22
1
2
1
2
1
2 2!
1
n
nk
n
s
k
n
sknk n
k
−
=
=

−+

−

=
+
∏
∑
.
Proposition 1.4. For all integer
1n≥
, there is the following identity:
( )
( )
( ) ( )
( )
12
22
1
22
2
0
0
2
12 2!
11
!1
nk
n
s
n
k
k
n
k
ks n
nk
k
nk
−
=
=
=


−
+

+


−=
+
 +


∏
∑∏
.
Proposition 1.5. For all integer
1n≥
, we have the following identity:
( )
( )
( ) ( )
( )
4
222
2
2
01
0
22
1!
2! 11
k
n
n
n
kr
k
nn
n
nk n
n kr kk
==
=

− 
−


−+ =

+
 +
∑∏∏
.
2. PROVING SOME NEW IDENTITIES IN COMBINATORIC BY USING THE
SYSTEMS OF LINEAR EQUATIONS
Example 2.1. Assume that 12
, , ... ,
n
αα α
∈
and
0, , 1, 2, ... ,
i
j ij n
α
+≠ = . Solve
the following system of linear equations:
12
11 1
12
22 2
12
... 1
12
... 1
12
...
... 1
12
n
n
n
nn n
xx x
n
xx x
n
xx x
n
αα α
αα α
αα α
+ ++ =
++ +
+ ++ =
++ +
+ ++ =
++ +
Proof: Consider
( ) ( )
( )
12
1
... 1
12
n
n
i
px
xx x
fx x x nx ix
=
= + + + −=
++ + +
∏
, where
( )
px
is a
polynomial of degree n. Since
( )
0
i
f
α
=
, therefore
( )
0
i
p
α
=
,
1, ..., in=
. In view of this
result, we get
( ) ( )( ) ( )
( )( ) ( )
12
...
1 2 ...
n
xx x
fx x x xn
αα α
−− −
= − ++ +
.
Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh
ISSN: 1844 – 9581 Mathematics Section
219
From
( ) ( )
( ) ( )
1
12
...
... 1
1 2 1 ...
n
n
xx
xx x
x x nx x xn
αα
−−
+ + + −=−
++ + + +
we deduce
( )( ) ( ) ( )( ) ( )
( )( )( ) ( ) ( ) ( )
( )( ) ( ) ( )( ) ( )
( )( ) ( )
12
34
12
2 3 ... 1 3 ...
1 2 4 ... 1 ...
...
1 2 ... 1 1 2 ...
...
n
n
xx x xn xx x xn
xx x x xn xx xn
xx x xn x x xn
xx x
αα α
+++++++
+ + + + ++ + +
+
+++ +−−++ +
=−− − −
From straightforward computation, we obtain the solution of the above system
equations
( ) ( )
( )
( ) ( )
( )
( ) ( )
( )
( ) ( )
( )
1
1
1
2
1
2
3
1
3
1
11
with 1
1!
12
with 2
1! 2 !
13
with 3
2! 3 !
...
1
with .
1!
−
=
−
=
−
=
−
=
−+
= = −
−
−+
= = −
−
−+
= = −
−
−+
= = −
−
∏
∏
∏
∏
n
n
i
i
n
n
i
i
n
n
i
i
n
nn
i
i
n
xx
n
xx
n
xx
n
n
x xn
n
α
α
α
α
฀
Proposition 2.2. Set
( ) ( )
1
n
i
i
xx
ϕα
=
= +
∏. Then, there are the following identities:
(i)
( ) ( ) ( )
0
1 1!
n
in
i
n
in
i
ϕ
=

−=−


∑.
(ii)
( ) ( )
0
1 1!
nin
n
i
nin
i
=

−=−


∑
.
(iii)
( ) ( )
0
11
n
in
i
n ni
ii
=
+

−=−


∑.
Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh
www.josa.ro Mathematics Section
Proof: (i) Similar to Example 2.1, we deduce that:
( ) ( )
( )( ) ( )
1
11
1! !
ni
n
i
i
xii ni
ϕ
−
=
−−
+− −
∑
( )( ) ( )
( )( ) ( )
12
...
1 2 ...
n
xx x
x x xn
αα α
−− −
= − ++ +. Substitute
0x=
, we have
( ) ( )
( ) ( ) ( )
1
1
1 10
1
!! !
ni n
n
i
i
in i n
ϕϕ
−+
=
−−
−=−
−
∑
or
( ) ( ) ( )
0
1 1!
n
in
i
n
in
i
ϕ
=

−=−


∑
(ii) Substitute
1
... 0
n
αα
= = =
in
( ) ( ) ( )
0
1 1!
n
in
i
n
in
i
ϕ
=

−=−


∑ we get
( ) ( )
0
1 1!
nin
n
i
nin
i
=

−=−


∑
.
(iii) Substitute
, 0, 1, ... ,
iii n
α
= =
we obtain
( ) ( )
0
11
n
in
i
n ni
ii
=
+

−=−


∑. □
Example 2.3. Solve the following system of linear equations:
22
1
1
1, ... , ; ,
=
+
=
+
=±± ∈
∑
n
kk
k
kk
sx y
sk s
s nxy ฀
Then, evaluate the sum
( )
( )( )
22
1
22 2
11
n
n
s
k
sk
sk
Sks k
=
=
≠
+
=
−+
∏
∑∏.
Proof: Consider
( ) ( )
( )
22
22
1
1
1
n
kk
n
k
k
px
xx y
fx x xk x xk
=
=
+
=−+ =
++
∑∏
, where
( )
px
is a
polynomial of the degree
2n≤
. Since
( )
0fs=
, we have
( )
0ps =
where
1, ... , sn=±±
.
We obtain
( )
( )( ) ( )
( )
22 2 2 2 2
22
1
1 2 ...
n
k
ax x x n
fx
x xk
=
−− −
=
+
∏
.
Since
( )( ) ( )
( )
22 2 2 2 2
22
22
1
1
1 2 ...
1n
kk
n
k
k
ax x x n
xx y
x xk x xk
=
=
−− −
+
−+ =
++
∑∏
we deduce
( ) ( )
(
)
( )(
) ( )
( )
( )( ) ( )
( )
( )( )
( )
( )
( )( ) ( )
22 22 2222 22
11
2
22 2 2 2 2 22 2 2 2
22
22 2 2 2 2
1 ... 2 3 ...
1 3 ... ... 1 2 ... 1
1 2 ...
nn
x xn xxxyx x xn
xxx y x x x n xxx y x x x n
ax x x n
−+ ++ + + + +
++ ++ ++++ ++ +−
=−− −
Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh
ISSN: 1844 – 9581 Mathematics Section
221
and obtain the solution
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
22
1
11
2
2
22
1
22
22
2
22
1
22
2
1 with 0
11
with
1
12
with 2
2
...
1
with
n
n
k
n
k
n
n
k
n
k
n
n
k
nn n
k
a nx
ak
x iy x i
k
ak
x iy x i
k
a nk
x iy x ni
kn
=
=
=
=
=
=
=−− =
−+
−+ = =
−
−+
−+ = =
−
−+
−+ = =
−
∏
∏
∏
∏
∏
∏
Hence
1 ... 0
n
yy= = =
and
( )
( )
22
1
22
1,
n
k
rn
k kr
rk
x
kr
=
= ≠
+
=
−
∏
∏
with
1, ... , rn=
.
Since
( )( ) ( )
( )
22 2 2 2 2
22
22
1
1
1 2 ...
1
x
n
k
n
k
k
ax x x n
xx
xk x xk
=
=
−− −
−+ =
++
∑∏
, we have
2
1
1
1
n
k
k
x
k
=
=
+
∑
or
( )
( )( )
22
1
22 2
1
1
1
n
n
s
n
k
sk
sk
ks k
=
=
≠
+
=
−+
∏
∑∏
by replacing
1x=
. □
Proposition 2.4. There is the following identity
( )
( )
( )
2 22
1
2
0
2
1
2 2!
1
n
nk
n
s
k
n
k sknk n
k
−
=
=

−+

−

=
+
∏
∑
.
Proof: Since
( )
( )( )
22
1
22 2
1
1
1
n
n
s
n
k
s nk
sk
ks k
=
=
≠
+
=
−+
∏
∑∏
by Example 2.3. and
( )
22
n
s nk
ks
≠
−=
∏
( )
n
s nk
ks
≠
−
∏
( )
n
s nk
ks
≠
+
∏
( )
( ) ( )
2
!!
1
2
nk nk nk
k
−
−+
= − , we have
( )
( ) ( ) ( )
( )
2 22
1
2
1
2
1
1 ! !1
n
n
s
nk
k
k sk
nk nk k
=
−
=
+
=
− − ++
∏
∑
.