Search results
Results from the WOW.Com Content Network
There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of n bits (digits 0 or 1) whose sum is k is given by (), while the number of ways to write = + + + where every a i is a nonnegative integer is ...
Hence the problem reduces to finding the binomial coefficient (). Also shown are the three corresponding 3-compositions of 4. The three-choose-two combination yields two results, depending on whether a bin is allowed to have zero items. In both results the number of bins is 3.
The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector space of dimension n over , a finite field with q elements; i.e. it is the number of points in the finite Grassmannian (,).
In mathematics, Pascal's rule (or Pascal's formula) is a combinatorial identity about binomial coefficients.It states that for positive natural numbers n and k, + = (), where () is a binomial coefficient; one interpretation of the coefficient of the x k term in the expansion of (1 + x) n.
In mathematics, Kummer's theorem is a formula for the exponent of the highest power of a prime number p that divides a given binomial coefficient. In other words, it gives the p-adic valuation of a binomial coefficient. The theorem is named after Ernst Kummer, who proved it in a paper, (Kummer 1852).
The two sets of three numbers which the Star of David theorem says have equal greatest common divisors also have equal products. [1] For example, again observing that the element 84 is surrounded by, in sequence, the elements 28, 56, 126, 210, 120, 36, and again taking alternating values, we have 28×126×120 = 2 6 ×3 3 ×5×7 2 = 56×210×36.
The Leibniz rule bears a strong resemblance to the binomial theorem, and in fact the binomial theorem can be proven directly from the Leibniz rule by taking () = and () =, which gives ( a + b ) n e ( a + b ) x = e ( a + b ) x ∑ k = 0 n ( n k ) a n − k b k , {\displaystyle (a+b)^{n}e^{(a+b)x}=e^{(a+b)x}\sum _{k=0}^{n}{\binom {n}{k}}a^{n-k}b ...
This expands the product into a sum of monomials of the form for some sequence of coefficients , only finitely many of which can be non-zero. The exponent of the term is n = ∑ i a i {\textstyle n=\sum ia_{i}} , and this sum can be interpreted as a representation of n {\displaystyle n} as a partition into a i {\displaystyle a_{i}} copies of ...