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In how many ways we can arrange k different items at n places such that every k item is at least at one place. n will always be greater than or equal to k.

eg.

When n = 2 and k = 1

We have (1,1)

Answer : 1

When n = 2 and k = 2

We have (1,2) and (2,1)

Answer : 2

When n = 3 and k = 2

We have (1,1,2),(1,2,1),(2,1,1),(1,2,2),(2,1,2) and (2,2,1)

Answer : 6

?

dbhrockzz
  • 31

2 Answers2

2

Let $S(i)$ be the set of functions from $n$ to $k$, where $i$ is not in the range.

Then $$ N(j)=\sum_{|A|=j}\left|\,\bigcap_{i\in A} S(i)\,\right|=\overbrace{\ \ \ \binom{k}{j}\ \ \ }^{\substack{\text{number of}\\\text{choices}\\\text{for }A}}\overbrace{\vphantom{\binom{k}{j}}\ (k-j)^n}^{\substack{\text{number of}\\\text{functions}\\\text{from $n\to k-j$}}}\tag{1} $$ The Generalized Inclusion-Exclusion Principle says that the number of functions in $0$ of the $S(i)$ is $$ \newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}} \begin{align} \sum_{j=0}^k(-1)^jN(j) &=\sum_{j=0}^k(-1)^j\binom{k}{j}(k-j)^n\\ &=\sum_{j=0}^k(-1)^{k-j}\binom{k}{j}j^n\\ &=\sum_{j=0}^k\sum_{i=0}^j(-1)^{k-j}\binom{k}{j}\binom{j}{i}\stirtwo{n}{i}i!\\ &=\sum_{i=0}^k\sum_{j=i}^k(-1)^{k-j}\binom{k}{i}\binom{k-i}{j-i}\stirtwo{n}{i}i!\\ &=\sum_{i=0}^k(-1)^{k-i}\binom{k}{i}[k=i]\stirtwo{n}{i}i!\\[6pt] &=\stirtwo{n}{k}k! \end{align} $$ Where $\stirtwo{n}{k}$ are Stirling Numbers of the Second Kind.

Examples from the question:
$\stirtwo{2}{1}1!=1$
$\stirtwo{2}{2}2!=2$
$\stirtwo{3}{2}2!=6$

robjohn
  • 345,667
  • Indeed we partition $n$ into $k$ non-empty sets of places as OP calls them, for ${n\brace k}$ possibilities. Now every set of places is assigned a value from the $k$ available ones with values being unique, producing a multiplier of $k!.$ (Consult introductory remarks of your Wikipedia link.) – Marko Riedel Jul 30 '17 at 22:51
1

Let $P(n,k)$ be the number of ways of doing this without the constraint that every number must appear at least once. Clearly $P(n,k) = k^n$.

Then note $P(n,k)$ contains $j^n$ arrangements containing only numbers from $1$ to $j$. So it might be tempting to say that there are $\binom{k}{j}j^n$ arrangements that only contain $j$ dfferent numbers or fewer however this is not the case. This is because if there are fewer than $j$ numbers used (say $l$) in the arrangement we can choose our $l$ numbers and then choose any other numbers we want and so each arrangement only using $l$ numbers is counted $\binom{k-l}{j-l}$ times.

Therefore $\sum_{l=1}^{j}\binom{k-l}{j-l}S(n,l) = \binom{k}{j}j^n$ where $S(n,l)$ is the number of arrangements using exactly $l$ different numbers. The answer to your question is $S(n,k)$.

Then by using inclusion exclusion we find $S(n,k) = \sum_{j=1}^{k}(-1)^{j+k}\binom{k}{j}j^n$.

Checking the $n=3,k=2$ case in your question gives $S(3,2) = \sum_{j=1}^{2}(-1)^{j}\binom{2}{j}j^3 = 8 - 2 = 6$ so it does indeed give the correct answer.

PJF49
  • 506