Question

I really need help ;-) suppose you have n items from which you choose r at a time. Explain why you must devide the number of permutations n!/(n-r)! by r! To find the number of combinations n!/r!(n-r)!*

Answer 1

Let's consider a simple example, choosing \(3\) objects from \(4\) (i.e. \(n=4\) and \(r=3\)). Let's call these items \(A\), \(B\), \(C\), and \(D\).

Here's a list of all the possible arrangements/permutations:
\[\begin{matrix}
\color{red}{ABC}&\color{red}{BAC}&\color{red}{CAB}&\color{blue}{DAB}\\
\color{blue}{ABD}&\color{blue}{BAD}&\color{green}{CAD}&\color{green}{DAC}\\
\color{red}{ACB}&\color{red}{BCA}&\color{red}{CBA}&\color{blue}{DBA}\\
\color{green}{ACD}&BCD&CBD&DBC\\
\color{blue}{ADB}&\color{blue}{BDA}&\color{green}{CDA}&\color{green}{DCA}\\
\color{green}{ADC}&BDC&CDB&DCB
\end{matrix}\]
There's a total here of \(P(4,3)=\dfrac{4!}{(4-3)!}=4!=24\) possible arrangements. The colored arrangements are all the permutations that contain the same letters. There are only \(4\) distinct permutations, \(C(4,3)=\dfrac{4!}{3!(4-3)!}=4\). We divided by \(3!\) here because each permutation has \(3!\) different ways of rearranging the same 3 letters.