− k + j j ) . {\displaystyle {\binom {n}{k}}{\binom {k}{j}}={\binom {n}{j}}{\binom {n-j}{k-j}}={\binom {n}{k-j}}{\binom {n-k+j}{j}}.} For constant n, we...

n k ) = n ( n − 1 ) ⋯ ( n − k + 1 ) k ( k − 1 ) ⋯ 1 , {\displaystyle {\binom {n}{k}}={\frac {n(n-1)\dotsb (n-k+1)}{k(k-1)\dotsb 1}},} which can be written...

p^{-r}=(1-q)^{-r}=\sum _{k=0}^{\infty }{\binom {-r}{{\phantom {-}}k}}(-q)^{k}=\sum _{k=0}^{\infty }{\binom {k+r-1}{k}}q^{k}} hence the terms of the probability...

f(k,n,p)=\Pr(X=k)={\binom {n}{k}}p^{k}(1-p)^{n-k}} for k = 0, 1, 2, ..., n, where ( n k ) = n ! k ! ( n − k ) ! {\displaystyle {\binom {n}{k}}={\frac {n...

n + 1 r + 1 ) . {\displaystyle {\binom {r}{r}}+{\binom {r+1}{r}}+{\binom {r+2}{r}}+\cdots +{\binom {n}{r}}={\binom {n+1}{r+1}}.} The name stems from...

binom {m}{1}}\sum _{2\leq a\leq n}x^{a}+{\binom {m}{2}}{\underset {ab\leq n}{\sum _{a=2}^{\infty }\sum _{b=2}^{\infty }}}x^{ab}+{\binom {m}{3}}{\underset...

) ( N n ) , {\displaystyle p_{X}(k)=\Pr(X=k)={\frac {{\binom {K}{k}}{\binom {N-K}{n-k}}}{\binom {N}{n}}},} where N {\displaystyle N} is the population...

the formula is symmetrical, ( n k ) = ( n n − k ) . {\textstyle {\binom {n}{k}}={\binom {n}{n-k}}.} A simple variant of the binomial formula is obtained...

s_{4B}(k)={\binom {2k}{k}}\sum _{j=0}^{k}4^{k-2j}{\binom {k}{2j}}{\binom {2j}{j}}^{2}={\binom {2k}{k}}\sum _{j=0}^{k}{\binom {k}{j}}{\binom {2k-2j}{k-j}}{\binom {2j}{j}}=1...

n^{m}={\frac {1}{m+1}}\left(B_{0}n^{m+1}-{\binom {m+1}{1}}B_{1}n^{m}+{\binom {m+1}{2}}B_{2}n^{m-1}-\cdots +(-1)^{m}{\binom {m+1}{m}}B_{m}n\right)} for all sums...

{\binom {m-1}{k-1}}{1}}\sum _{n=m}^{\infty }{1 \over {\binom {n}{k}}}\\&={\frac {\binom {m-1}{k-1}}{1}}\cdot {\frac {k}{k-1}}\cdot {\frac {1}{\binom...

cups chosen, there are ( 8 4 ) = 8 ! 4 ! ( 8 − 4 ) ! = 70 {\displaystyle {\binom {8}{4}}={\frac {8!}{4!(8-4)!}}=70} possible combinations. The frequencies...

The Gaussian binomial coefficient, written as ( n k ) q {\displaystyle {\binom {n}{k}}_{q}} or [ n k ] q {\displaystyle {\begin{bmatrix}n\\k\end{bmatrix}}_{q}}...

+ ⋯ + ( c 2 2 ) + ( c 1 1 ) {\displaystyle N={\binom {c_{k}}{k}}+\cdots +{\binom {c_{2}}{2}}+{\binom {c_{1}}{1}}} . The fact that a unique sequence corresponds...

_{k=0}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=1}^{n+1}{\binom {n}{k-1}}f^{(n+1-k)}g^{(k)}\\&={\binom {n}{0}}f^{(n+1)}g^{(0)}+\sum _{k=1}^{n}{\binom...

{\displaystyle {\frac {{\binom {a+c}{a}}{\binom {b+d}{b}}(a+b)!(c+d)!}{n!}}={\frac {{\binom {a+c}{a}}{\binom {b+d}{b}}}{\binom {n}{a+b}}}} Another derivation:...

k − i m n − 1 ) {\displaystyle {\binom {n}{k}}_{m-1}=\sum _{i=0}^{\lfloor k/m\rfloor }(-1)^{i}{\binom {n}{i}}{\binom {n-1+k-im}{n-1}}} Multinomial distribution...

Pascal's rule determines that ( n k ) = n ! k ! ( n − k ) ! , {\displaystyle {\binom {n}{k}}={\frac {n!}{k!(n-k)!}},} for all integers 0 ≤ k ≤ n. In this sense...

0=(1-1)^{t}={\binom {t}{0}}-{\binom {t}{1}}+{\binom {t}{2}}-\cdots +(-1)^{t}{\binom {t}{t}}.} Using the fact that ( t 0 ) = 1 {\displaystyle {\binom {t}{0}}=1}...

≡ ∏ i = 0 k ( m i n i ) ( mod p ) , {\displaystyle {\binom {m}{n}}\equiv \prod _{i=0}^{k}{\binom {m_{i}}{n_{i}}}{\pmod {p}},} where m = m k p k + m k...

}{\binom {1}{m}}{\binom {1}{n-m}}\\&=p^{n}\left(1-p\right)^{2-n}\left[{\binom {1}{0}}{\binom {1}{n}}+{\binom {1}{1}}{\binom {1}{n-1}}\right]\\&={\binom...

in the OEIS) The central binomial coefficient ( 2 n n ) {\displaystyle {\binom {2n}{n}}} is the number of arrangements where there are an equal number...

_{i}{\binom {a_{i}}{2}}+\sum _{j}{\binom {b_{j}}{2}}\right]-\left[\sum _{i}{\binom {a_{i}}{2}}\sum _{j}{\binom {b_{j}}{2}}\right]\right/{\binom {n}{2}}}}}...

&=\sum _{k=0}^{n-1}{\binom {n}{k+1}}(2k-1)!!(2n-2k-3)!!\\&=\sum _{k=1}^{n}{\binom {n}{k}}(2k-3)!!(2(n-k)-1)!!\\&=\sum _{k=0}^{n}{\binom {2n-k-1}{k-1}}{\frac...

\right)={\binom {n+k-1}{k-1}}={\binom {10+4-1}{4-1}}={\binom {13}{3}}=286,} where the multiset coefficient ( ( k n ) ) {\displaystyle \left(\!\!{\binom {k}{n}}\...

{n}{1}}={\binom {n+0}{1}}={\binom {n}{1}}} (linear numbers), P 2 ( n ) = n ( n + 1 ) 2 = ( n + 1 2 ) {\displaystyle P_{2}(n)={\frac {n(n+1)}{2}}={\binom {n+1}{2}}}...

_{j=0}^{k}{\binom {j+m}{j}}{\binom {n-m-j}{k-j}}&=\sum _{j=0}^{k}(-1)^{j}{\binom {-m-1}{j}}(-1)^{k-j}{\binom {m+1+k-n-2}{k-j}}\\&=(-1)^{k}\sum _{j=0}^{k}{\binom {-m-1}{j}}{\binom...

− 2 n x + ( n 2 ) , {\displaystyle {\mathcal {K}}_{2}(x;n)=2x^{2}-2nx+{\binom {n}{2}},} K 3 ( x ; n ) = − 4 3 x 3 + 2 n x 2 − ( n 2 − n + 2 3 ) x + (...

getting a Yarborough is ( 32 13 ) ( 52 13 ) {\displaystyle {\frac {\binom {32}{13}}{\binom {52}{13}}}} which is 347 , 373 , 600 635 , 013 , 559 , 600 {\displaystyle...

b ) / 2 ⌋ . {\displaystyle {\binom {p(a,b)}{q(a,b)}}={\begin{cases}{\binom {1}{b}},&{\text{if }}b=a+1{\text{,}}\\{\binom {p(a,m)q(m,b)+p(m,b)}{q(a,m)q(m...