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- where <math>\binom{n}{k} = \frac{n!}{k!(n-k)!}</math> is a [[binomial coefficient]]. In other words, the coefficients when <math>(a + b)^n</math> is expande ...-m</math> terms of the product must contribute a <math>b</math>. Thus, the coefficient of <math>a^m b^{n-m}</math> is the number of ways to choose <math>m</math>5 KB (935 words) - 12:11, 20 February 2024
- By the [[Multinomial Theorem]], the summands can be written as ...me in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:8 KB (1,332 words) - 16:37, 17 September 2023
- Alternatively, we can use a multinomial coefficient to obtain the answer directly: <math>\binom{8}{3,2,3}=\frac{8!}{3!\cdot2!\c3 KB (491 words) - 03:24, 4 November 2022
- The '''Multinomial Theorem''' states that where <math>\binom{n}{j_1; j_2; \ldots ; j_k}</math> is the [[multinomial coefficient]] <math>\binom{n}{j_1; j_2; \ldots ; j_k}=\dfrac{n!}{j_1!\cdot j_2!\cdots j3 KB (476 words) - 18:37, 4 January 2023
- ...ng function <cmath>\left(x^3+x^4+x^6\right)^{10}</cmath> where we want the coefficient of <math>x^{41}.</math> ...>{x^3}^{10}\left(1+x+x^3\right)^{10}</cmath> and then see that we want the coefficient of <math>x^{11}</math> in <math>\left(1+x+x^3\right)^{10}.</math>6 KB (909 words) - 14:39, 8 August 2022
- ...fficient of <math>a^2b^2</math> is <math>2\binom{6}{3,2,1}=120</math>. The coefficient of <math>ab</math> is <math>\binom{6}{2,2,1,1}+2\binom{6}{3,2,1}=300</math>11 KB (1,677 words) - 22:54, 4 February 2022
- We will use casework similar to problem 64 but using multinomial coefficients (instead of binomial coefficients) to account The Multinomial Coefficient <math>\binom{(a+b+c)}{a,b,c}</math> is defined as <math>\frac{(a+b+c)!}{a!b2 KB (271 words) - 04:21, 28 January 2019
- ...lternatively there is no factor of 5 in the denominator of the multinomial coefficient unless an element is repeated 5 times). Therefore to evaluate the answer mo14 KB (2,244 words) - 16:08, 31 October 2024