二項式係數可排列成帕斯卡三角形 在數學 上，二項式係數 是二項式定理 中各項的係數 。一般而言，二項式係數由兩個非負整數
n
{\displaystyle n}
k
{\displaystyle k}
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
(
1
+
x
)
n
{\displaystyle (1+x)^{n}}
x
k
{\displaystyle x^{k}}
係數 ，因此一定是非負整數。如果將二項式係數
(
n
0
)
,
(
n
1
)
,
…
,
(
n
n
)
{\displaystyle {\binom {n}{0}},{\binom {n}{1}},\dots ,{\binom {n}{n}}}
n
=
0
,
1
,
2
,
…
{\displaystyle n=0,1,2,\dots }
帕斯卡三角形 。
二項式係數常見於各數學領域中，尤其是組合數學 。事實上，
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
n
{\displaystyle n}
k
{\displaystyle k}
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
n
{\displaystyle n}
k
{\displaystyle k}
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
n
{\displaystyle n}
複數 的情況，而且仍然被稱為二項式係數。
雖然二項式係數在西元10世紀就已經被發現（見帕斯卡三角形 ），但表達式
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
安德烈亞斯·馮·厄廷格豪森 首次始用[ 注 1] Halayudha 印度教 典籍《賓伽羅 的計量聖典》（chandaḥśāstra）。約1150年，印度數學家婆什迦羅第二 於其著作《Lilavati 》[ 注 2]
二項式係數亦有不同的符號 表達方式，包括：
C
(
n
,
k
)
{\displaystyle C(n,k)}
n
C
k
{\displaystyle _{n}C_{k}}
n
C
k
{\displaystyle ^{n}C_{k}}
C
n
k
{\displaystyle C_{n}^{k}}
C
k
n
{\displaystyle C_{k}^{n}}
[ 注 3] 置換 數
P
k
n
{\displaystyle P_{k}^{n}}
n , k )。
對於非負整數
n
{\displaystyle n}
k
{\displaystyle k}
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
(
1
+
x
)
n
{\displaystyle (1+x)^{n}}
x
k
{\displaystyle x^{k}}
係數 ，即
(
1
+
x
)
n
=
∑
k
=
0
n
(
n
k
)
x
k
=
(
n
0
)
+
(
n
1
)
x
+
⋯
+
(
n
n
)
x
n
{\displaystyle (1+x)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}={\binom {n}{0}}+{\binom {n}{1}}x+\cdots +{\binom {n}{n}}x^{n}}
事實上，若
x
{\displaystyle x}
y
{\displaystyle y}
交換環 上的元素，則
(
x
+
y
)
n
=
∑
k
=
0
n
(
n
k
)
x
n
−
k
y
k
{\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{n-k}y^{k}}
此數的另一出處在組合數學，表達了從
n
{\displaystyle n}
k
{\displaystyle k}
n
{\displaystyle n}
k
{\displaystyle k}
(
1
+
X
)
n
{\displaystyle (1+X)^{n}}
n
{\displaystyle n}
i
{\displaystyle i}
n
{\displaystyle n}
k
{\displaystyle k}
X
k
{\displaystyle X^{k}}
n
{\displaystyle n}
k
{\displaystyle k}
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
n
{\displaystyle n}
位元 （如數字0或1）組成的所有序列中，其和為
k
{\displaystyle k}
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
k
=
a
1
+
a
2
+
⋯
+
a
n
{\displaystyle k=a_{1}+a_{2}+\cdots +a_{n}}
a
i
{\displaystyle a_{i}}
(
n
+
k
−
1
k
)
{\displaystyle {\tbinom {n+k-1}{k}}}
k
{\displaystyle k}
除展開二項式或點算組合數量之外，尚有多種方式計算
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
以下遞歸 公式可計算二項式係數：
(
n
k
)
=
(
n
−
1
k
−
1
)
+
(
n
−
1
k
)
∀
n
,
k
∈
N
{\displaystyle {\binom {n}{k}}={\binom {n-1}{k-1}}+{\binom {n-1}{k}}\quad \forall n,k\in \mathbb {N} }
其中特別指定：
(
n
0
)
=
1
∀
n
∈
N
∪
{
0
}
,
{\displaystyle {\binom {n}{0}}=1\quad \forall n\in \mathbb {N} \cup \{0\},}
(
0
k
)
=
0
∀
k
∈
N
.
{\displaystyle {\binom {0}{k}}=0\quad \forall k\in \mathbb {N} .}
此公式可由計算
(
1
+
x
)
n
−
1
(
1
+
x
)
{\displaystyle (1+x)^{n-1}(1+x)}
x
k
{\displaystyle x^{k}}
{
1
,
2
,
⋯
,
n
}
{\displaystyle \left\{1,2,\cdots ,n\right\}}
k
{\displaystyle k}
n
{\displaystyle n}
n
{\displaystyle n}
顯然，如果
k
>
n
{\displaystyle k>n}
(
n
k
)
=
0
{\displaystyle {\tbinom {n}{k}}=0}
n
{\displaystyle n}
(
n
n
)
=
1
{\displaystyle {\tbinom {n}{n}}=1}
帕斯卡三角形 。
個別二項式係數可用以下公式計算：
(
n
k
)
=
n
k
_
k
!
=
n
(
n
−
1
)
(
n
−
2
)
⋯
[
n
−
(
k
−
1
)
]
k
(
k
−
1
)
(
k
−
2
)
⋯
1
=
∏
i
=
1
k
n
−
(
k
−
i
)
i
,
{\displaystyle {\binom {n}{k}}={\frac {n^{\underline {k}}}{k!}}={\frac {n(n-1)(n-2)\cdots [n-(k-1)]}{k(k-1)(k-2)\cdots 1}}=\prod _{i=1}^{k}{\frac {n-(k-i)}{i}},}
上式中第一個分數的分子是一階乘冪 。此公式可以二項式係數在計算組合數量的意義理解：分子為從
n
{\displaystyle n}
k
{\displaystyle k}
k
{\displaystyle k}
二項式係數最簡潔的表達式是階乘 :
(
n
k
)
=
n
!
k
!
(
n
−
k
)
!
for
0
≤
k
≤
n
.
{\displaystyle {\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}\quad {\mbox{for }}\ 0\leq k\leq n.}
其中「
n
!
{\displaystyle n!}
n
{\displaystyle n}
(
n
−
k
)
!
{\displaystyle (n-k)!}
(
n
k
)
=
(
n
n
−
k
)
for
0
≤
k
≤
n
.
{\displaystyle {\binom {n}{k}}={\binom {n}{n-k}}\quad {\mbox{for }}\ 0\leq k\leq n.}
1
若將
n
{\displaystyle n}
α
{\displaystyle \alpha }
逆元素 的交換環 中的一元素，則二項式係數可籍乘數公式擴展[ 注 4]
(
α
k
)
=
α
k
_
k
!
=
α
(
α
−
1
)
(
α
−
2
)
⋯
(
α
−
k
+
1
)
k
(
k
−
1
)
(
k
−
2
)
⋯
1
for
k
∈
N
and arbitrary
α
.
{\displaystyle {\binom {\alpha }{k}}={\frac {\alpha ^{\underline {k}}}{k!}}={\frac {\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -k+1)}{k(k-1)(k-2)\cdots 1}}\quad {\mbox{for }}k\in \mathbb {N} {\mbox{ and arbitrary }}\alpha .}
此定義能使二項式公式一般化（其中一單項為1），故
(
α
k
)
{\displaystyle {\tbinom {\alpha }{k}}}
(
1
+
X
)
α
=
∑
k
=
0
∞
(
α
k
)
X
k
.
{\displaystyle (1+X)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}X^{k}.}
2
此公式對任何複數
α
{\displaystyle \alpha }
X
{\displaystyle X}
|
X
|
<
1
{\displaystyle \left\vert X\right\vert <1}
X
{\displaystyle X}
冪級數 的恆等式，即係數為常數1，任意冪之級數定義，且在此定義下，對於冪的恆等式成立，例如
(
1
+
X
)
α
(
1
+
X
)
β
=
(
1
+
X
)
α
+
β
and
(
(
1
+
X
)
α
)
β
=
(
1
+
X
)
α
β
.
{\displaystyle (1+X)^{\alpha }(1+X)^{\beta }=(1+X)^{\alpha +\beta }\quad {\mbox{and}}\quad ((1+X)^{\alpha })^{\beta }=(1+X)^{\alpha \beta }.}
若
α
{\displaystyle \alpha }
n
{\displaystyle n}
k
>
n
{\displaystyle k>n}
α
{\displaystyle \alpha }
帕斯卡三角形的第1000行，垂直排列，且以灰階表示係數的十進制數位，向右對齊，故左邊邊界約是二項式係數的對數，圖中可見數族形成一對數凹數列 帕斯卡法則 是一重要的遞歸 等式：
(
n
k
)
+
(
n
k
+
1
)
=
(
n
+
1
k
+
1
)
,
{\displaystyle {n \choose k}+{n \choose k+1}={n+1 \choose k+1},}
3
此式可以用於數學歸納法 ，以証明
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
n
{\displaystyle n}
k
{\displaystyle k}
k
!
{\displaystyle k!}
k
{\displaystyle k}
公式(1) 中得出。
帕斯卡法則建構出帕斯卡三角形 :
0:
1
1:
1
1
2:
1
2
1
3:
1
3
3
1
4:
1
4
6
4
1
5:
1
5
10
10
5
1
6:
1
6
15
20
15
6
1
7:
1
7
21
35
35
21
7
1
8:
1
8
28
56
70
56
28
8
1
第
n
{\displaystyle n}
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
k
=
0
,
…
,
n
{\displaystyle k=0,\ldots ,n}
(
x
+
y
)
5
=
1
x
5
+
5
x
4
y
+
10
x
3
y
2
+
10
x
2
y
3
+
5
x
y
4
+
1
y
5
{\displaystyle (x+y)^{5}={\boldsymbol {1}}x^{5}+{\boldsymbol {5}}x^{4}y+{\boldsymbol {10}}x^{3}y^{2}+{\boldsymbol {10}}x^{2}y^{3}+{\boldsymbol {5}}xy^{4}+{\boldsymbol {1}}y^{5}}
在斜線上相鄰項的差就是上一斜線上的數值，此乃上述遞歸等式(3
二項式係數是組合數學 中的重要課題，因其可用於眾多常見的點算問題中，例如
共有
(
n
k
)
{\displaystyle {\tbinom {n}{k}}}
n
{\displaystyle n}
k
{\displaystyle k}
組合 。
共有
(
n
+
k
−
1
k
)
{\displaystyle {\tbinom {n+k-1}{k}}}
n
{\displaystyle n}
k
{\displaystyle k}
多重集 。
共有
(
n
+
k
k
)
{\displaystyle {\tbinom {n+k}{k}}}
字符串 包含
k
{\displaystyle k}
n
{\displaystyle n}
共有
(
n
+
1
k
)
{\displaystyle {\tbinom {n+1}{k}}}
k
{\displaystyle k}
n
{\displaystyle n}
[ 參 1]
卡塔蘭數 是
(
2
n
n
)
n
+
1
{\displaystyle {\frac {\tbinom {2n}{n}}{n+1}}}
統計學 中的二項式分佈 是
(
n
k
)
p
k
(
1
−
p
)
n
−
k
{\displaystyle {\tbinom {n}{k}}p^{k}(1-p)^{n-k}\!}
貝茲曲線 的公式。就任就非負整數
k
{\displaystyle k}
(
t
k
)
{\displaystyle \scriptstyle {\binom {t}{k}}}
k
!
{\displaystyle k!}
(
t
k
)
=
(
t
)
k
k
!
=
(
t
)
k
(
k
)
k
=
t
(
t
−
1
)
(
t
−
2
)
⋯
(
t
−
k
+
1
)
k
(
k
−
1
)
(
k
−
2
)
⋯
(
2
)
(
1
)
;
{\displaystyle {\binom {t}{k}}={\frac {(t)_{k}}{k!}}={\frac {(t)_{k}}{(k)_{k}}}={\frac {t(t-1)(t-2)\cdots (t-k+1)}{k(k-1)(k-2)\cdots (2)(1)}};\,\!}
此為帶有理數 係數，變量是
t
{\displaystyle t}
多項式 ，可對任意實數或複數
t
{\displaystyle t}
牛頓廣義二項式定理 。
就任意
k
{\displaystyle k}
(
t
k
)
{\displaystyle {\tbinom {t}{k}}}
k
{\displaystyle k}
p
(
t
)
{\displaystyle p(t)}
p
(
0
)
=
p
(
1
)
=
…
=
p
(
k
−
1
)
=
0
{\displaystyle p(0)=p(1)=\ldots =p(k-1)=0}
p
(
k
)
=
1
{\displaystyle p(k)=1}
其係數可以第一類斯特靈數 表示，即：
(
t
k
)
=
∑
i
=
0
k
s
k
,
i
k
!
t
i
{\displaystyle {\binom {t}{k}}=\sum _{i=0}^{k}{\frac {s_{k,i}}{k!}}t^{i}}
(
t
k
)
{\displaystyle {\tbinom {t}{k}}}
導數 可以對數微分 計算：
d
d
t
(
t
k
)
=
(
t
k
)
∑
i
=
0
k
−
1
1
t
−
i
.
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\binom {t}{k}}={\binom {t}{k}}\sum _{i=0}^{k-1}{\frac {1}{t-i}}\,.}
在任何包含Q 域 中，最多
d
{\displaystyle d}
∑
k
=
0
d
a
k
(
t
k
)
{\displaystyle \sum _{k=0}^{d}a_{k}{\binom {t}{k}}}
a
k
{\displaystyle a_{k}}
p
(
0
)
,
p
(
1
)
,
…
,
p
(
k
)
{\displaystyle p(0),p(1),\ldots ,p(k)}
第k 差分 ，亦即：
[ 注 5]
a
k
=
∑
i
=
0
k
(
−
1
)
k
−
i
(
k
i
)
p
(
i
)
.
{\displaystyle a_{k}=\sum _{i=0}^{k}(-1)^{k-i}{\binom {k}{i}}p(i).}
3.5
每一多項式
(
t
k
)
{\displaystyle {\tbinom {t}{k}}}
k
{\displaystyle k}
帕斯卡法則 以歸納法証明）。故此，二項式係數多項式的整數線性組合亦為整數值。反之，(3.5
k
{\displaystyle k}
R
{\displaystyle R}
K
[
t
]
{\displaystyle K[t]}
R
{\displaystyle R}
R
{\displaystyle R}
整數值多項式
3
t
(
3
t
+
1
)
2
{\displaystyle {\frac {3t(3t+1)}{2}}}
9
(
t
2
)
+
6
(
t
1
)
+
0
(
t
0
)
{\displaystyle 9{\tbinom {t}{2}}+6{\tbinom {t}{1}}+0{\tbinom {t}{0}}}
從
t
=
1
,
2
,
3
{\displaystyle t=1,2,3}
3
t
(
3
t
+
1
)
2
=
6
,
21
,
45
{\displaystyle {\frac {3t(3t+1)}{2}}=6,21,45}
帕斯卡矩陣 的逆 可算出：
(
(
t
−
1
0
)
(
t
−
1
1
)
(
t
−
1
2
)
)
(
1
0
0
−
1
1
0
1
−
2
1
)
(
6
21
45
)
=
(
(
t
−
1
0
)
(
t
−
1
1
)
(
t
−
1
2
)
)
(
6
15
9
)
{\displaystyle {\begin{pmatrix}{\tbinom {t-1}{0}}&{\tbinom {t-1}{1}}&{\tbinom {t-1}{2}}\end{pmatrix}}{\begin{pmatrix}1&0&0\\-1&1&0\\1&-2&1\end{pmatrix}}{\begin{pmatrix}6\\21\\45\end{pmatrix}}={\begin{pmatrix}{\tbinom {t-1}{0}}&{\tbinom {t-1}{1}}&{\tbinom {t-1}{2}}\end{pmatrix}}{\begin{pmatrix}6\\15\\9\end{pmatrix}}}
=
6
(
t
−
1
0
)
+
15
(
t
−
1
1
)
+
9
(
t
−
1
2
)
=
6
(
t
1
)
+
9
(
t
2
)
{\displaystyle =6{\tbinom {t-1}{0}}+15{\tbinom {t-1}{1}}+9{\tbinom {t-1}{2}}=6{\tbinom {t}{1}}+9{\tbinom {t}{2}}}
這種二項式係數多項式結合朱世傑恆等式 應用於等冪求和 。
階乘公式能聯繫相鄰的二項式係數，例如在
k
{\displaystyle k}
n
{\displaystyle n}
(
n
+
1
k
)
=
(
n
k
)
+
(
n
k
−
1
)
{\displaystyle {\binom {n+1}{k}}={\binom {n}{k}}+{\binom {n}{k-1}}}
(
n
k
)
=
n
k
(
n
−
1
k
−
1
)
{\displaystyle {\binom {n}{k}}={\frac {n}{k}}{\binom {n-1}{k-1}}}
(
n
−
1
k
)
−
(
n
−
1
k
−
1
)
=
n
−
2
k
n
(
n
k
)
.
{\displaystyle {\binom {n-1}{k}}-{\binom {n-1}{k-1}}={\frac {n-2k}{n}}{\binom {n}{k}}.}
兩個組合數相乘可作變換：
(
n
i
)
(
i
m
)
=
(
n
m
)
(
n
−
m
i
−
m
)
{\displaystyle {\binom {n}{i}}{\binom {i}{m}}={\binom {n}{m}}{\binom {n-m}{i-m}}}
[ 參 2]
∑
r
=
0
n
(
n
r
)
=
2
n
{\displaystyle \sum _{r=0}^{n}{\binom {n}{r}}=2^{n}}
∑
r
=
0
k
(
n
+
r
−
1
r
)
=
(
n
+
k
k
)
{\displaystyle \sum _{r=0}^{k}{\binom {n+r-1}{r}}={\binom {n+k}{k}}}
∑
r
=
0
n
−
k
(
−
1
)
r
(
n
+
1
)
k
+
r
+
1
(
n
−
k
r
)
=
(
n
k
)
−
1
{\displaystyle \sum _{r=0}^{n-k}{\frac {(-1)^{r}(n+1)}{k+r+1}}{\binom {n-k}{r}}={\binom {n}{k}}^{-1}}
∑
r
=
0
n
(
d
n
d
r
)
=
1
d
∑
r
=
1
d
(
1
+
e
2
π
r
i
d
)
d
n
{\displaystyle \sum _{r=0}^{n}{\binom {dn}{dr}}={\frac {1}{d}}\sum _{r=1}^{d}(1+e^{\frac {2\pi ri}{d}})^{dn}}
[ 參 3]
∑
i
=
m
n
(
a
+
i
i
)
=
(
a
+
n
+
1
n
)
−
(
a
+
m
m
−
1
)
{\displaystyle \sum _{i=m}^{n}{\binom {a+i}{i}}={\binom {a+n+1}{n}}-{\binom {a+m}{m-1}}}
(
a
+
m
m
−
1
)
+
(
a
+
m
m
)
+
(
a
+
m
+
1
m
+
1
)
+
.
.
.
+
(
a
+
n
n
)
=
(
a
+
n
+
1
n
)
{\displaystyle {\binom {a+m}{m-1}}+{\binom {a+m}{m}}+{\binom {a+m+1}{m+1}}+...+{\binom {a+n}{n}}={\binom {a+n+1}{n}}}
F
n
=
∑
i
=
0
∞
(
n
−
i
i
)
{\displaystyle F_{n}=\sum _{i=0}^{\infty }{\binom {n-i}{i}}}
[ 參 4]
F
n
−
1
+
F
n
=
∑
i
=
0
∞
(
n
−
1
−
i
i
)
+
∑
i
=
0
∞
(
n
−
i
i
)
=
1
+
∑
i
=
1
∞
(
n
−
i
i
−
1
)
+
∑
i
=
1
∞
(
n
−
i
i
)
=
1
+
∑
i
=
1
∞
(
n
+
1
−
i
i
)
=
∑
i
=
0
∞
(
n
+
1
−
i
i
)
=
F
n
+
1
{\displaystyle F_{n-1}+F_{n}=\sum _{i=0}^{\infty }{\binom {n-1-i}{i}}+\sum _{i=0}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n-i}{i-1}}+\sum _{i=1}^{\infty }{\binom {n-i}{i}}=1+\sum _{i=1}^{\infty }{\binom {n+1-i}{i}}=\sum _{i=0}^{\infty }{\binom {n+1-i}{i}}=F_{n+1}}
∑
i
=
m
n
(
i
a
)
=
(
n
+
1
a
+
1
)
−
(
m
a
+
1
)
{\displaystyle \sum _{i=m}^{n}{\binom {i}{a}}={\binom {n+1}{a+1}}-{\binom {m}{a+1}}}
(
m
a
+
1
)
+
(
m
a
)
+
(
m
+
1
a
)
.
.
.
+
(
n
a
)
=
(
n
+
1
a
+
1
)
{\displaystyle {\binom {m}{a+1}}+{\binom {m}{a}}+{\binom {m+1}{a}}...+{\binom {n}{a}}={\binom {n+1}{a+1}}}
∑
r
=
0
n
(
n
r
)
2
=
(
2
n
n
)
{\displaystyle \sum _{r=0}^{n}{\binom {n}{r}}^{2}={\binom {2n}{n}}}
∑
i
=
0
n
(
r
1
+
n
−
1
−
i
r
1
−
1
)
(
r
2
+
i
−
1
r
2
−
1
)
=
(
r
1
+
r
2
+
n
−
1
r
1
+
r
2
−
1
)
{\displaystyle \sum _{i=0}^{n}{\binom {r_{1}+n-1-i}{r_{1}-1}}{\binom {r_{2}+i-1}{r_{2}-1}}={\binom {r_{1}+r_{2}+n-1}{r_{1}+r_{2}-1}}}
[ 參 5]
(
1
−
x
)
−
r
1
(
1
−
x
)
−
r
2
=
(
1
−
x
)
−
r
1
−
r
2
{\displaystyle (1-x)^{-r_{1}}(1-x)^{-r_{2}}=(1-x)^{-r_{1}-r_{2}}}
(
1
−
x
)
−
r
1
(
1
−
x
)
−
r
2
=
(
∑
n
=
0
∞
(
r
1
+
n
−
1
r
1
−
1
)
x
n
)
(
∑
n
=
0
∞
(
r
2
+
n
−
1
r
2
−
1
)
x
n
)
=
∑
n
=
0
∞
(
∑
i
=
0
n
(
r
1
+
n
−
1
−
i
r
1
−
1
)
(
r
2
+
i
−
1
r
2
−
1
)
)
x
n
{\displaystyle (1-x)^{-r_{1}}(1-x)^{-r_{2}}=(\sum _{n=0}^{\infty }{\binom {r_{1}+n-1}{r_{1}-1}}x^{n})(\sum _{n=0}^{\infty }{\binom {r_{2}+n-1}{r_{2}-1}}x^{n})=\sum _{n=0}^{\infty }(\sum _{i=0}^{n}{\binom {r_{1}+n-1-i}{r_{1}-1}}{\binom {r_{2}+i-1}{r_{2}-1}})x^{n}}
(
1
−
x
)
−
r
1
−
r
2
=
∑
n
=
0
∞
(
r
1
+
r
2
+
n
−
1
r
1
+
r
2
−
1
)
x
n
{\displaystyle (1-x)^{-r_{1}-r_{2}}=\sum _{n=0}^{\infty }{\binom {r_{1}+r_{2}+n-1}{r_{1}+r_{2}-1}}x^{n}}
∑
i
=
0
k
(
n
i
)
(
m
k
−
i
)
=
(
n
+
m
k
)
{\displaystyle \sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}={\binom {n+m}{k}}}
(
n
+
k
k
)
2
=
∑
j
=
0
k
(
k
j
)
2
(
n
+
2
k
−
j
2
k
)
{\displaystyle {\binom {n+k}{k}}^{2}=\sum _{j=0}^{k}{\binom {k}{j}}^{2}{\binom {n+2k-j}{2k}}}
Benjamin, Arthur T. ; Quinn, Jennifer (2003). Proofs that Really Count: The Art of Combinatorial Proof （頁面存檔備份 ，存於網際網路檔案館 ）, Mathematical Association of America.Bryant, Victor . Aspects of combinatorics . Cambridge University Press. 1993. ISBN 0521419743 Flum, Jörg; Grohe, Martin. Parameterized Complexity Theory . Springer. 2006 [2011-07-28 ] . ISBN 978-3-540-29952-3存檔 於2007-11-18）. Fowler, David . The Binomial Coefficient Function . The American Mathematical Monthly (Mathematical Association of America). January 1996, 103 (1): 1–17. JSTOR 2975209 doi:10.2307/2975209 Graham, Ronald L. ; Knuth, Donald E. ; Patashnik, Oren . Concrete Mathematics Second. Addison-Wesley. 1994: 153 –256. ISBN 0-201-55802-5 Higham, Nicholas J. Handbook of writing for the mathematical sciences . SIAM . 1998: 25 . ISBN 0898714206 Knuth, Donald E. The Art of Computer Programming, Volume 1: Fundamental Algorithms ISBN 0-201-89683-4 Singmaster, David . Notes on binomial coefficients. III. Any integer divides almost all binomial coefficients. J. London Math. Soc. (2). 1974, 8 (3): 555–560. doi:10.1112/jlms/s2-8.3.555 Shilov, G. E. Linear algebra . Dover Publications. 1977. ISBN 9780486635187 
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