1- have two sets: set A containing $k$ numbers and set B containing $m$ numbers. want to build a set c whose elements are a pair $(i,j)$ where $i$ belongs to A and $j$ belongs to B. Permutation says the set C would contain $\frac{(k+m)!}{k!m!}$ how could I avoid repetition? or does the above formula exclude repetition? 2- what if the set C is to build using only set A, i.e. the pair $(i,j)$ from set C both $i$ and $j$ belongs to A. Is the number of elements in the set C equal to $\frac{(k+k)!}{k!k!}$ i.e. would this avoid repetition?
Asked
Active
Viewed 1,269 times
0
-
Sorry, why wouldn't $C$ just contain $k\times m$ pairs? – lulu Sep 05 '17 at 17:30
-
But $C=A\times B$ (where $\times$ is the cartesian product) and has $km$ elements. – ajotatxe Sep 05 '17 at 17:30
-
1Can you clarify your question? As it stands, it doesn't really make sense. – lulu Sep 05 '17 at 17:34
1 Answers
0
The set $C$ has $km$ elements. $\frac{(k+m)!}{k!m!}$ is the number of unordered pairs $\{i,j\}$ where $i,j\in A\cup B$ and $i\neq j$. And that assuming that $A\cap B=\emptyset$.
For example
ajotatxe
- 65,084
-
OK, I see my confusion. But how about the second case where both sets are identical? how could repetition be avoided? it is certainly less than $kXk$, correct? – bijan karimi Sep 05 '17 at 18:21