See the following recursion tree for n = 5 an k = 2. )^{-1} \equiv ((x-1)! {\displaystyle (x+y)} + y y brightness_4 x x Quite often you come across the problem of computing binomial coefficients modulo some $m$. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Bell Numbers (Number of ways to Partition a Set), Find minimum number of coins that make a given value, Greedy Algorithm to find Minimum number of Coins, K Centers Problem | Set 1 (Greedy Approximate Algorithm), Minimum Number of Platforms Required for a Railway/Bus Station, K’th Smallest/Largest Element in Unsorted Array | Set 1, K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time), K’th Smallest/Largest Element in Unsorted Array | Set 3 (Worst Case Linear Time), k largest(or smallest) elements in an array | added Min Heap method, Top 20 Dynamic Programming Interview Questions, Space and time efficient Binomial Coefficient, http://www.csl.mtu.edu/cs4321/www/Lectures/Lecture%2015%20-%20Dynamic%20Programming%20Binomial%20Coefficients.htm, Sum of product of r and rth Binomial Coefficient (r * nCr), Eggs dropping puzzle (Binomial Coefficient and Binary Search Solution), Fibonomial coefficient and Fibonomial triangle, Replace the maximum element in the array by coefficient of range, Mathematics | PnC and Binomial Coefficients, Middle term in the binomial expansion series, Find sum of even index binomial coefficients, Program to print binomial expansion series, Sum of product of consecutive Binomial Coefficients, Add two numbers without using arithmetic operators, Travelling Salesman Problem | Set 1 (Naive and Dynamic Programming), Write a program to print all permutations of a given string, Set in C++ Standard Template Library (STL), Write Interview x b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. . y y All combinations of v, returned as a matrix of the same type as v. 1 Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Likning (7a) gjelder for alle verdier av m, mens likning (7b) gjelder for alle verdier av j. {\displaystyle y^{3}} = x + . e It is easy to deduce this using the analytic formula. e (C står for det engelske ordet combination) og leses «n over k». ) . y {\displaystyle x+y} n We often say "n choose k" when referring to the binomial coefficient. }{k! C'est la base de calcul du nombre de combinaisons de k éléments parmi n. Exemple : Le nombre de combinaisons au loto est de 5 parmi 49 soit $ {49 \choose 5} = 1906884 $ combinaisons possibles. m , . It is believed that this formula, as well as the triangle which allows efficient calculation of the coefficients, was discovered by Blaise Pascal in the 17th century. 3 {\displaystyle (x+y)(x+y)} x Finally, in some situations it is beneficial to precompute all the factorials in order to produce any necessary binomial coefficient with only two divisions later. ; likeså for y, slik at vi får leddet Binomial Coefficients Recursion tree for C(5,2). For example, given a group of 15 footballers, there is exactly \\( \binom {15}{11} = 1365\\) ways we can form a football team. x Denne metoden gjør det mulig å raskt regne ut binomial koeffisienter uten å måtte bruke brøk eller multiplikasjon. Den angir hvor mange forskjellige kombinasjoner en kan velge Analytic formulafor the calculation: (nk)=n!k!(n−k)! + Dette gjentakelsesforholdet kan brukes til å bevise, ved matematisk induksjon, at C(n, k) er et naturlig tall for alle n og k, et faktum som ikke er umiddelbart tydelig ut ifra definisjonen. The advantage of this method is that intermediate results never exceed the answer and calculating each new table element requires only one addition. ( {\displaystyle k} Pascals regel er det viktige gjentakelsesforholdet. $$ \binom n k = \frac n k \binom {n-1} {k-1} $$, Sum over $k$: x Therefore, we can replace our fraction with a product $k$ fractions, each of which is real-valued. ) {\displaystyle y^{4}} C — All combinations of v matrix. + Writing code in comment? ( x Nevertheless we can compute the binomial coefficient. For example: og Proof From the definition, $${n \choose k} = \dfrac{n! y ) Det første leddet får vi ved å gange x fra begge faktorene, slik at vi får This can be advantageous when using long arithmetic, when the memory does not allow precomputation of the whole Pascal's triangle. z k That is because \\( \binom {n} {k} \\) is equal to the number of distinct ways \\(k\\) items can be picked from n items. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^k. After precomputing all values for $g$ and $c$, which can be done efficiently using dynamic programming in $\mathcal{O}(n)$, we can compute the binomial coefficient in $O(\log m)$ time. y | 2 fra n faktorer av ( Fra (2), etter å ha derivert og satt inn x = y = 1. The left-Hand side represents the value of the current iteration which will be obtained by this statement. z y p n $$ 1 \binom n 1 + 2 \binom n 2 + \cdots + n \binom n n = n 2^{n-1} $$. Here are the simplest of them: The first, straightforward formula is very easy to code, but this method is likely to overflow even for relatively small values of $n$ and $k$ (even if the answer completely fit into some datatype, the calculation of the intermediate factorials can lead to overflow). ( First, let's count the number of ordered selections of $k$ elements. y Binomialkoeffisienten av n og k blir også skrevet som C(n, k), nCk eller permutasjoner. But if $p \le \max(k, n-k)$, then at least one of $k!$ and $(n-k)!$ are not coprime with $m$, and therefore we cannot compute the inverses - they don't exist. {\displaystyle C_{n}^{k}} {\displaystyle x^{n-k}y^{k}} 1) Optimal Substructure The value of C(n, k) can be recursively calculated using the following standard formula for Binomial Coefficients. {\displaystyle {z \choose k}} ( ganget med x og , som former ledd som følger. Following is the Top-down approach of dynamic programming to finding the value of the Binomial Coefficient. )^{-1} \mod m.$$. eller {\displaystyle x^{2}} + {\displaystyle (x_{1}+...+x_{m})^{n}} n x This formula can be easily deduced from the problem of ordered arrangement (number of ways to select $k$ different elements from $n$ different elements). y {\displaystyle x^{2}y} Then we can write the binomial coefficient as: ) 2 That is because \\( \binom {n} {k} \\) is equal to the number of distinct ways \\(k\\) items can be picked from n items. y $$\binom n k \equiv n! . x Perhaps it was discovered by a Persian scholar Omar Khayyam. And let $g(x) := \frac{x!}{p^{c(x)}}$. Vi teller mulighetene ved å betrakte de n! . ( Når eksponenten er 1, blir ) kan defineres for alle komplekse tall z og alle naturlige tall k som følger: Denne generaliseringen er kjent som den generelle binomialkoeffisienten og er brukt i utredningen av binomialformelen og oppfyller egenskapene (3) og (7). {\displaystyle x^{2}y^{2}} A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. y ( close, link Or precompute all inverses and all powers of $p$, and then compute the binomial coefficient in $O(1)$. Siden C(n, k) er definert som null hvis k > n, er summen endelig. 2 ( Leddene i utvidelsen av . That is because \\( \binom {n} {k} \\) is equal to the number of distinct ways \\(k\\) items can be picked from n items. + 2 The idea is the following: + Binomial coefficients have many different properties. + The following formula holds: $${n \choose 0} = 1,$$ where ${n \choose 0}$ denotes the binomial coefficient. k Den enkle binomialkoeffisienten er tilfellet der m=2. $$ {\binom n 0}^2 + {\binom n 1}^2 + \cdots + {\binom n n}^2 = \binom {2n} n $$, Weighted sum: 1 n ( The Problem Write a function that takes two parameters n and k and returns the value of Binomial Coefficient C(n, k). har formen, hvor Thanks to AK for suggesting this method. n! y . This formula can be easily deduced from the problem of ordered arrangement (number of ways to select k different elements from n different elements). permutasjoner, og de k faktorene av y har k! For å spare plass bruker vi den første av disse tre notasjonene. {\displaystyle n} Dette er opprinnelsen til Pascals trekant, som er diskutert nedenfor. ) Ved å utvide (1+x)m (1+x)n-m = (1+x)n med (2). generalization of Lucas's theorem for prime powers, Codeforces - Points, Lines and Ready-made Titles, Symmetry rule: elementer i en mengde med Men xy leddet kan formes av x fra den første og y fra den andre faktoren, eller y fra den første og x fra den andre faktoren; derfor får leddet koeffisienten 2. {\displaystyle x^{3}} ) Time Complexity: O(n*k) Auxiliary Space: O(n*k)Following is a space-optimized version of the above code. . 2 ( Sett fra et annet perspektiv, for å forme Nevertheless, it was known to the Chinese mathematician Yang Hui, who lived in the 13th century. Else we compute the value and store in the lookup table. 1 Diskusjonen kan videreføres til tilfellet hvor hver faktor er en sum av flere variabler, som naturligvis leder til definisjonen av en multinomialkoeffisient. n {\displaystyle xy^{2}} References: http://www.csl.mtu.edu/cs4321/www/Lectures/Lecture%2015%20-%20Dynamic%20Programming%20Binomial%20Coefficients.htmPlease write comments if you find anything incorrect, or you want to share more information about the topic discussed above. Når eksponenten er 2, blir Likning (7a) er Vandermonde's konvolusjonsformel (etter Alexandre-Théophile Vandermonde) og er essensielt en form for Chu-Vandermonde identiteten. That is because \\( \binom {n} {k} \\) is equal to the number of distinct ways \\(k\\) items can be picked from n items. j For eksempel , og dette viser seg i den numeriske "symmetrien" i Pascals trekant. {\displaystyle (x+y)^{n}} Let the prime factorization of $m$ be $m = p_1^{e_1} p_2^{e_2} \cdots p_h^{e_h}$. When $n$ is too large, the $\mathcal{O}(n)$ algorithms discussed above become impractical. y Binomialkoeffisienten x First, let's count the number of ordered selections of k elements. y 4 {\displaystyle k/p^{j}} Time Complexity: O(n*k) Auxiliary Space: O(k)Explanation: 1==========>> n = 0, C(0,0) = 1 1–1========>> n = 1, C(1,0) = 1, C(1,1) = 1 1–2–1======>> n = 2, C(2,0) = 1, C(2,1) = 2, C(2,2) = 1 1–3–3–1====>> n = 3, C(3,0) = 1, C(3,1) = 3, C(3,2) = 3, C(3,3)=1 1–4–6–4–1==>> n = 4, C(4,0) = 1, C(4,1) = 4, C(4,2) = 6, C(4,3)=4, C(4,4)=1 So here every loop on i, builds i’th row of pascal triangle, using (i-1)th rowAt any time, every element of array C will have some value (ZERO or more) and in next iteration, value for those elements comes from previous iteration. k Use this step-by-step solver to calculate the binomial coefficient. LOI.BINOMIALE. Hver permutasjon fremstilles som en uordnet liste med tall fra 1 til n. Vi velger en x fra de n−k første faktorene, og en y fra de resterende k faktorene; på denne måten vil hver permutasjon bidra til leddet , hvor vi allerede vet at fakultetet av n. Ifølge Nicholas J. Higham, ble denne notasjonen introdusert av Albert von Ettinghausen i 1826, selv om disse tallene var kjent i århundrer før dette; se Pascals trekant. If $p > \max(k, n-k)$, then we can use the same method as described in the previous section. 4 . elementer. Binomial coefficient, returned as a nonnegative scalar value. {\displaystyle (x+y)^{2}=x^{2}+2xy+y^{2}} 01:40. x m + ) Memoization Approach : The idea is to create a lookup table and follow the recursive top-down approach. BC, got similar results. ) er begge − ganget med y, slik at leddets koeffisienten blir 3+3. ( ) n For example, given a group of 15 footballers, there is exactly \\( \binom {15}{11} = 1365\\) ways we can form a football team. n til (derav navnet): Dette generaliseres ved binomialformelen, som tillater eksponenten n å være et komplekst tall, (spesielt tillater dette n å være ethvert reelt tall, ikke nødvendigvis bare positive heltall). As a result, we get the formula of the number of ordered arrangements: $n (n-1) (n-2) \cdots (n - k + 1) = \frac {n!} \cdot (k! $p^c ~|~ x!$. {\displaystyle m} y , For example, given a group of 15 footballers, there is exactly \\( \binom {15}{11} = 1365\\) ways we can form a football team. {\displaystyle (x+y)^{1}} Likning (7a) generaliserer likning (3). }` Differansene mellom elementer på andre diagonaler er elementene på forrige diagonal – slik som følger av gjentakelsesforholdet (3) ovenfor. ( {\displaystyle (x+y)^{3}} Recurrence formula (which is associated with the famous "Pascal's Triangle"): $$ \binom n k = \binom {n-1} {k-1} + \binom {n-1} k $$. C++ implementation: Here we carefully cast the floating point number to an integer, taking into account that due to the accumulated errors, it may be slightly less than the true value (for example, $2.99999$ instead of $3$). k ) y ) Binomialkoeffisienten har en q-analog generalisering kjent som Gauss-binomet. Please use ide.geeksforgeeks.org, generate link and share the link here. For example: e Like other typical Dynamic Programming(DP) problems, re-computations of the same subproblems can be avoided by constructing a temporary 2D-array C[][] in a bottom-up manner. The previously discussed approach of Pascal's triangle can be used to calculate all values of $\binom{n}{k} \bmod m$ for reasonably small $n$, since it requires time complexity $\mathcal{O}(n^2)$. Now we compute the binomial coefficient modulo some arbitrary modulus $m$. x {(n-k)!}$. ( = $$\binom n k = \frac {n!} For eksempel, ved å se på den femte raden i trekanten, kan en straks lese av at. C++ implementation: If the entire table of values is not necessary, storing only two last rows of it is sufficient (current $n$-th row and the previous $n-1$-th). Dette utledes fra (2) ved at The interesting thing is, that $g(x)$ is now free from the prime divisor $p$. Therefore $g(x)$ is coprime to m, and we can compute the modular inverses of $g(k)$ and $g(n-k)$. The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^k. {\displaystyle n} y + x på et unikt vis. Binomialkoeffisienter er av stor betydning i kombinatorikk, fordi de gir ferdige formler for visse hyppige telleproblemer: Binomialkoeffisienter forekommer også i formelen for binomisk distribusjon i statistikk og i formelen for en Bézier kurve. 2 Le coefficient binomial est noté, ), da oppstår leddet på to måter, fra tilstøtende koeffisienter med sammenlagt grad 3. Dette er det samme som å si at summen av elementene av en rad i Pascals trekant alltid tilsvarer to opphøyd i et heltall. antall virkelig distinkte måter å forme leddet This gives us $h$ different congruences. Når eksponenten er 3, reduseres )^{-1} \cdot ((n-k)! }$ by $k!$. {\displaystyle x^{4}} / {\displaystyle y^{2}} + . \\( (a+1)^n= \binom {n} {0} a^n+ \binom {n} {1} + a^n-1+...+ \binom {n} {n} a^n \\) . Binomial coefficients are also the coefficients in the expansion of $(a + b) ^ n$ (so-called binomial theorem): $$ (a+b)^n = \binom n 0 a^n + \binom n 1 a^{n-1} b + \binom n 2 a^{n-2} b^2 + \cdots + \binom n k a^{n-k} b^k + \cdots + \binom n n b^n $$. k E So the Binomial Coefficient problem has both properties (see this and this) of a dynamic programming problem. x {(n-k)! 1 When $m$ is not square-free, a generalization of Lucas's theorem for prime powers can be applied instead of Lucas's theorem. ( {\displaystyle x^{n-k}y^{k}} x And afterwards we can compute the binomial coefficient in $O(\log m)$ time. représente factorielle n soit, Binomialkoeffisienten er en grunnleggende matematisk funksjon i det matematiske delområdet kombinatorikk. x z n As a result, we get the formula of the number of ordered arrangements: n(n−1)(n−2)⋯(n−k+1)=n!(n−k)!. , $$\binom n k = \frac {g(n) p^{c(n)}} {g(k) p^{c(k)} g(n-k) p^{c(n-k)}} = \frac {g(n)} {g(k) g(n-k)}p^{c(n) - c(k) - c(n-k)}$$. On les note () (lu « k parmi n » ) ou C k n (lu « combinaison de k parmi n »). j Derfor er n!/(n−k)!k! When the modulo $m$ is prime, there are 2 options: When $m$ is not prime but square-free, the prime factors of $m$ can be obtained and the coefficient modulo each prime factor can be calculated using either of the above methods, and the overall answer can be obtained by the Chinese Remainder Theorem. Dette kan bevises ved induksjon av n ved å bruke (3). y x f Binomial coefficient is an integer that appears in the [binomial expansion] (/show/calculator/binomial-theorem). ) x x = {\displaystyle (x+y)^{2}(x+y)} Dette er viktig i teorien om differenslikninger og kan bli sett på som en diskret analog til Taylors teorem. By using the recurrence relation we can construct a table of binomial coefficients (Pascal's triangle) and take the result from it. Binomialkoeffisienten av et naturlig tall n og et heltall k er definert som det naturlige tallet. Igjen oppstår ekstremene Moreover, Indian mathematician Pingala, who lived earlier in the 3rd. {\displaystyle n} 2 )^{-1} \cdot x^{-1}$ and the method for computing all inverses in $O(n)$. Since the same subproblems are called again, this problem has Overlapping Subproblems property. k Avez-vous des suggestions pour améliorer cette page . Dette foreslår en induksjon. x Denne artikkelen bruker materiale fra følgende PlanetMath artikler, som er lisensiert under GFDL: Formler som inneholder binomialkoeffisienter, https://no.wikipedia.org/w/index.php?title=Binomialkoeffisient&oldid=20922490, Artikler med autoritetsdatalenker fra Wikidata, Creative Commons-lisensen Navngivelse-Del på samme vilkår. , kjent som et multi-indeks. . $$ \binom n k = \binom n {n-k} $$, Factoring in: k Following is a simple recursive implementation that simply follows the recursive structure mentioned above. For large values of n, there will be many common subproblems. ) k y $$ \sum_{k = 0}^n \binom n k = 2 ^ n $$, Sum over $n$: + permutasjonene av faktorene. que l’on prononce « k parmi n » ou « combinaison de k parmi n »), donne donc le nombre de parties de k éléments dans un ensemble total de n éléments, avec k ≤ n, (ce qui revient à dire que le coefficient binomial est le nombre de chemins conduisant à k succès). For eksempel, listen ⟨4,1,2,3⟩ velger x fra faktorene 4 og 1, og y velger faktorer fra 2 og 3, som en måte å forme leddet Le coefficient binomial, dit "k parmi n" ou "combinaison de k parmi n" pour n, un entier naturel et k entier naturel inférieur ou égal à n, est le nombre de sous-ensembles de k éléments dans un ensemble de n éléments. },$$ x For solving binomial coefficients we have use from formula $\\frac{n!}{k!(n-k)! rad nummer n inneholder tallene C(n, k) for k = 0,...,n. Den konstrueres ved å begynne med enere på utsiden og så legge sammen nabotall og skrive summen rett under. . Les deux notations sont préconisées par la norme ISO/CEI 80000-2:2009 [1] : la première est celle du « coefficient binomial » (2-10.4) et la seconde celle du « nombre de combinaisons sans répétition » (2-10.6). y A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. og som er null når k < 0 eller k > n. Her betegner n! $m = p^b$ for some prime $p$. We can easily move to unordered arrangements, noting that each unordered arrangement corresponds to exactly $k!$ ordered arrangements ($k!$ is the number of possible permutations of $k$ elements). For eksempel kan binomialkoeffisienten brukes til å beregne hvor mange mulige tallkombinasjoner som finnes i Lotto: Hvilket igjen betyr at sannsynligheten for å få sju rette i Lotto på en gitt rekke er: Binomialkoeffisientene er koeffisienter i utvidelsen av binomet En gunstig notasjon bruker en liste av variabler 2 ) 2 er større enn brøkdelen av Her betegner F(n + 1) Fibonacci-tallene. = p By using our site, you m edit The following code only uses O(k). so if we want to compute it modulo some prime $m > n$ we get Note that for $n \lt k$ the value of $\binom n k$ is assumed to be zero. x ) Experience. 2 Use this step-by-step solver to calculate the binomial coefficient. I Precious Mirror of the Four Elements (1303), nevner Zhu Shijie trekanten som en eldgammel metode for å løse binomialkoeffisienter, noe som indikerer at metoden var kjent for kinesiske matematikere fem århundrer før Pascal. Therefore, this method often can only be used with long arithmetic: Note that in the above implementation numerator and denominator have the same number of factors ($k$), each of which is greater than or equal to 1. Calcule les probabilités pour une distribution binomiale. For example, given a group of 15 footballers, there is exactly \\( \binom {15}{11} = 1365\\) ways we can form a football team. `n! ) . Primtallsdivisorer til C(n, k) kan tolkes som følger: Hvis p er et primtall og r er den høyeste eksponenten slik at slik at x Before computing any value, we check if it is already in the lookup table. {\displaystyle p^{r}} Here we want to compute the binomial coefficient modulo some prime power, i.e. Attention reader! This approach can handle any modulo, since only addition operations are used. {\displaystyle (x+y)^{n}=(y+x)^{n}} $$ \sum_{m = 0}^n \binom m k = \binom {n + 1} {k + 1} $$, Sum over $n$ and $k$: er delelig med C(n, k), da er r likt antallet naturlige tall j slik at brøkdelen av + {\displaystyle \mathbf {x} =(x_{1},...,x_{m})} objekter fra en mengde av The formula for the binomial coefficients is E We often say "n choose k" when referring to the binomial coefficient. Det er ikke vanskelig å vise at rekkens konvergensradius er 1. 2 {\displaystyle xy^{2}} Following is Dynamic Programming based implementation. x n We even can compute the binomial coefficient in $O(1)$ time if we precompute the inverses of all factorials in $O(\text{MAXN} \log m)$ using the regular method for computing the inverse, or even in $O(\text{MAXN})$ time using the congruence $(x! x For example, your function should return 6 for n = 4 and k = 2, and it should return 10 for n = 5 and k = 2. + , må vi velge y fra k av faktorene og x fra resten. / og $$ \sum_{k = 0}^m \binom {n + k} k = \binom {n + m + 1} m $$, Sum of the squares: In statement, C[j] = C[j] + C[j-1] The right-hand side represents the value coming from the previous iteration (A row of Pascal’s triangle depends on the previous row). Men den distinkte listen ⟨1,4,3,2⟩ gjør akkurat det samme utvalget; formelen for binomialkoeffisienten må fjerne denne overflødigheten. {\displaystyle k} | We get the final formula by dividing $\frac {n!} Outil pour générer les combinaisons. ulike objekter (uten tilbakelegging, uavhengig av rekkefølgen). Binomial coefficients $\binom n k$ are the number of ways to select a set of $k$ elements from $n$ different elements without taking into account the order of arrangement of these elements (i.e., the number of unordered sets). En mathématiques, un choix de k objets parmi n objets discernables, ou l'ordre n'intervient pas, se représente par ensemble d'éléments, dont le cardinal est le coefficient binomial. − x . . x ( Le coefficient binomial est utilisé principalement dans les calculs de dénombrements et de probabilités. x Slik at for eksponenten 4, har hvert ledd sammenlagt grad (sum av eksponentene) på 4, med 4-k faktorer av x og k faktorer av y. Hvis k ikke er 0 eller 1 (leddene x k ( x The following are the common definitions of Binomial Coefficients. , og koeffisienten til et slikt ledd er multinomialkoeffisienten. De n-k faktorene av x har (n−k)! The time complexity can be considered to be $\mathcal{O}(n^2)$. n som følger direkte fra definisjonen. n k Fra (2) ved å bruke at x = y = 1. . y See this for Space and time efficient Binomial Coefficient Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. Sagt på en annen måte angir binomialkoeffisienten antall delmengder med {\displaystyle x^{2}} We can easily … n y The merit of the Newton is that he generalized this formula for exponents that are not natural. er enten 2xy ganget med x eller {\displaystyle n/p^{j}} ganget med y, som gir koeffisienten 3; likeså oppstår . There are $n$ ways to select the first element, $n-1$ ways to select the second element, $n-2$ ways to select the third element, and so on. Notice, if $c(n) - c(k) - c(n-k) \ge b$, than $p^b ~|~ p^{c(n) - c(k) - c(n-k)}$, and the binomial coefficient is $0$. Let $c(x)$ be that number. + r x Denne metoden gjør det mulig å raskt regne ut binomial koeffisienter uten å måtte bruke brøk eller multiplikasjon.
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