23.3: Applications
If \(vert A \vert = n\text{,}\) then the number of ways to partition \(A\) into \(m\) disjoint subsets \(A_1, A_2, \ldots, A_m\text{,}\) with each subset of predetermined size \(\vert A_j \vert = i_j\text{,}\) is
\begin{equation*} \binom{n}{i_1,i_2,\ldots,i_m} \text{.} \end{equation*}
- Proof Idea.
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There are \(C^n_{i_1}\) possibilities for \(A_1\text{.}\) After choosing \(A_1\text{,}\) there are \(C^{n - i_1}_{i_2}\) possibilities for \(A_2\text{.}\) After choosing \(A_2\text{,}\) there are \(C^{n - i_1 - i_2}_{i_3}\) possibilities for \(A_2\text{.}\) Continue in this fashion, all the way to \(A_m\text{,}\) then multiply all the combination formula expressions together.
- Alternative Proof Idea.
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Going back to basic counting principles, we can approach this in the same way that we came up with the factorial formula for the choose function. Choosing a permutation of \(A\) (\(n!\) ways) gives us an instance of the desired partition of \(A\) by setting \(A_1\) to be the subset consisting of the first \(i_1\) objects in the permutation, then setting \(A_2\) to be the subset consisting of the next \(i_2\) objects in the permutation, and so on. However, the ordering of the elements inside any such subset \(A_j\) does not matter, and we would get the same partition if we took our permutation of \(A\) and again permuted the “clusters” corresponding to each subset \(A_j\text{.}\) Since there are \(i_j!\) ways to permute subset \(A_j\text{,}\) we should divide \(n!\) by each of the factorials \(i_j!\text{.}\)
In the above theorem, the order \(A_1, A_2, \ldots, A_m\) matters!
Suppose \(x_1, x_2, \ldots, x_m\) are distinct letters in the alphabet \(\Sigma\text{.}\) For \(i_1 + i_2 + \cdots + i_m = n\text{,}\) the number of words in \(\Sigma ^{\ast}\) of length \(n\) which consist of exactly \(i_1\) \(x_1\)'s, \(i_2\) \(x_2\)'s, \(\ldots\text{,}\) and \(i_m\) \(x_m\)'s is the multinomial coefficient
\begin{equation*} \binom{n}{i_1,i_2,\ldots,i_m} \text{.} \end{equation*}
- Proof Idea.
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If we view each letter \(x_i\) as a variable and each word made up of the letters \(x_1,\ldots,x_m\) as a product of these variables, then each of the words we want to count gives us one way to achieve a term of \(x_1^{i_1} \cdots x_m^{i_m}\) in the expansion of \((x_1 + \cdots + x_m)^n\text{.}\) The number of such ways is the multinomial coefficient.
How many different \(9\)-digit integers can we form from three \(3\)s, four \(6\)s and two \(9\)s?
Solution
The number of integers of the desired digit composition is the multinomial coefficient
\begin{equation*} \binom{9}{3,4,2} = \dfrac{9!}{3!4!2!} = \dfrac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 2} = 9 \cdot 4 \cdot 7 \cdot 5 = 1,260\text{.} \end{equation*}