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Multinomial Coefficients
Let X be a set of n elements. Suppose that we have two colors of paint, say red and blue, and we are going to choose a subset of k elements to be painted red with the rest painted blue.
Multinomial Coefficients
Let \(X\) be a set of \(n\) elements. Suppose that we have two colors of paint, say red and blue, and we are going to choose a subset of \(k\) elements to be painted red with the rest painted blue. Then the number of different ways this can be done is just the binomial coefficient \(\binom{n}{k}\). Now suppose that we have three different colors, say red, blue, and green. We will choose \(k_1\) to be colored red, \(k_2\) to be colored blue, and the remaining \(k_3 = n - (k_1+k_2)\) are to be colored green. We may compute the number of ways to do this by first choosing \(k_1\) of the \(n\) elements to paint red, then from the remaining \(n-k_1\) elements choosing \(k_2\) to paint blue, and then painting the remaining \(k_3\) elements green. It is easy to see that the number of ways to do this is \[\begin{aligned}\end{aligned}\] Numbers of this form are called multinomial coefficients; they are an obvious generalization of the binomial coefficients. The general notation is: \[\begin{aligned}\end{aligned}\] \(\binom{n}{k_1,k_2,k_3,\dots,k_r}\) multinomial coefficient
For example, \[\begin{aligned}\end{aligned}\]
Note that there is some overkill in this notation, since the value of \(k_r\) is determined by \(n\) and the values for \(k_1\), \(k_2,\dots,k_{r-1}\). For example, with the ordinary binomial coefficients, we just write \(\binom{8}{3}\) and not \(\binom{8}{3,5}\).
Example
How many different rearrangements of the string: \[\begin{aligned}\end{aligned}\] are possible if all letters and characters must be used?
Solution
To answer this question, we note that there are a total of \(45\) characters distributed as follows: 3A's, 1C, 1D, 7E's, 1G, 1H, 4I's, 1K, 5L's, 2M's, 2N's, 1O, 4R's, 2S's, 6T's, 1U, 1W, and 2!'s. So the number of rearrangements is \[\begin{aligned}\end{aligned}\]
Just as with binomial coefficients and the Binomial Theorem, the multinomial coefficients arise in the expansion of powers of a multinomial:
Condensed: the full section is in Keller & Trotter, Applied Combinatorics.
Issa inti L-ebda kalkulatur ma jista' jsolvi din il-problema, iżda l-biċċiet tagħha jistgħu jiġu kkalkulati. Ipprova waħda minn hawn taħt, jew ittajpja tiegħek stess.
Prattika (1)
Ipprova kull waħda fuq il-karta l-ewwel.Iżvela t-tweġiba biex tiċċekkja; dawk ivverifikati jistgħu jinfetħu fis-solver għal kull pass.
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Look again at the trinomial in . What would be the coefficient on \(x^{30}y^{80}z^{10}\) when this is expanded? Explain your reasoning.
Kont b'xejn iżid noti fuq kull lezzjoni, rekord ta' dak li lest, problemi solvuti tiegħek f'post wieħed, u għalliem inti tista' tistaqsi dwar din il-paġna.Il-matematika nnifisha hija miftuħa għal kulħadd, log fil jew le.
Irreġistra issa LoginSimboli użati hawnhekk
Taptap kwalunkwe simbolu għad-definizzjoni sħiħa, stampa, u dak li kull ittra fiha tfisser.
Mistoqsijiet li n-nies jistaqsu
Permutation or combination?
Ask whether order matters. A lock code is a permutation (order matters); a hand of cards is a combination (it does not).
What is a graph in this sense?
Dots (vertices) joined by lines (edges), not a plot. Road maps, social networks and molecules are graphs; questions like "is there a route" and "how few colours" are graph theory.
Partijiet minn din il-paġna huma adattati minn Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Kondensat u mill-ġdid spjegat hawn; żbalji huma tagħna.
Aktar fil Combinatorics & Graph Theory
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