23.3: Applications
- Page ID
- 83519
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
-
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.
-
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.
-
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*}