Hey Developer’s, I’m back with a new topic which is Multinomial Coefficients in the series of statistics foundations.

Quick Refresher

A binomial coefficient equals the number of combinations of k items that can be selected from a set of n items.

_nC_k=(n; k)=(n!)/((n-k)!k!),

So, let’s get started …

Multinomial Coefficients

It’s not just of separating two group of elements but actually separating several groups ,more than 2

Example: 10 students need to from 3 groups consisting of 4,3 and 3 members repectively.How many ways can students be assigned to these groups ?

Solution = First Group : Choose 4 students out of 10, 10C4 arrangements.

We are left with 6 students: number of ways to split them is 6C3

(10C4)(6C3) = (10!/4!6!) * (6!/3!3!) = 10!/4!3!3! = 4200

Multinomial Coefficient written as

{n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}

Just like the binomial coefficients,

{\displaystyle (x_{1}+x_{2}+\cdots +x_{m})^{n}=\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}\prod _{t=1}^{m}x_{t}^{k_{t}}\,,}

Example 1: Picking 4,3,3 students out of 10 : (10 C 4,3,3) = 10!/4!3!3! = 4200

Example 2: Number of ways to arrange 3 a’s, 4 b’s and 5 c’s is : (12 C 3,4,5) = 12!/3!4!5! = 27,720

Next Post will be on Probability of a union of events

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