Probability of getting at least one pair in a combination Assume that I have 6 symbols which I label $\{x_1,x_2,x_3,y_1,y_2,y_3\}$. Now I make a random 6-symbol  combination with repetition, such that every combination has the same probability of occuring. For example we could have drawn 2 times $x_1$, 3 times $y_2$ and 1 time $y_3$, but we could have also drawn 5 times $x_1$ and 1 time $x_2$ equally likely.
The question that I am now interested in is given a random combination with repetition as described above, what is the probability that the combination has at least one $x$ and one $y$ with the same index?
Supplementary: What is the probability if we consider a $M$-symbol combination made from a set of symbols $\{x_1,...,x_N,y_1,...,y_N\}$?
Edit: It is a problem that I have come up with myself. It is related to a statistical physics problem where the combinations are my microscopic realizations.
I tried to tackle it in the following way:
The total number of possible combinations with repetition is $\left(\binom{6}{6}\right)=\binom{11}{6}=462$.
Then I thought the easiest way was to calculate the combinations with no $x$,$y$ pair with the same indices present. Since there may not be any pairs present we need to choose a symbol with index 1, a symbol with index 2 and a symbol with index 3 and make with them a 6-symbol word. We can choose $2^3=8$ of such triples of symbols with each $\left(\binom{3}{6}\right)=\binom{8}{6}=28$ combinations. Wich would give $8\cdot28=224$ combinations or a probability of $p=1-224/462\approx0.515$. Which is isn't correct because of some extra combinations that I have counted multiple times. However I have no clue how to correct for these extra countings, in a structured way.
So I also tried the other approach of directly counting the probability of at least one pair. There are 3 possible pairs. Assume we have drawn the pair $x_1,y_1$, this has $\left(\binom{6}{4}\right)=\binom{9}{4}=126$ combinations. Then we need the amount of combinations that are not already counted and contain the pair $x_2$,$y_2$, naively I could say it is 2$\left(\binom{5}{4}\right)=2\binom{8}{4}$, because we can't have a pair with index 1. But again my naive counting mechanism have problems with multiple countings, which I do not know how to efficiently deal with.
I have tried to simulate it such that I could check my answers but none of them are correct. The simulation gives for N=3 and M=6: $p\approx 0.68$.
 A: As far as i have understood , the selection order of elements does not matter ,so we only care about the combination.
Note to OP: To make calculations easier , i will use generating functions , and i assume that you know about it.
For your example , there are $462$ different combinations using the set of $\{x_1,x_2,x_3,y_1,y_2,y_3\}$. I have calculated it using generating functions such that $$[a^6]\bigg(\frac{1}{1-a}\bigg)^6=462$$
Then , our denominator is $462$. Because of it is stated in original question , each combination have $1/462$ probability to occur.
Now , we want the cases that  "the combination has at least one x and one y with the same index". In other words , the combination contains at least one $\{x_1,y_1\}$ or at least one $\{x_2,y_2\}$ or at least one $\{x_3,y_3\}$.
I think that the connective "or" reminds you P.I.E (inclusion-exclusion).Then,

*

*The number of combinations contain $\{x_i,y_i\}$ : $$[a^6]\bigg(\frac{a}{1-a}\bigg)^2\bigg(\frac{1}{1-a}\bigg)^4=126$$


*The number of combinations contain $\{x_i,y_i\}$ and $\{x_k,y_k\}$ : $$[a^6]\bigg(\frac{a}{1-a}\bigg)^4\bigg(\frac{1}{1-a}\bigg)^2=21$$


*The number of combinations contain $\{x_i,y_i\}$ , $\{x_k,y_k\}$ ,and $\{x_z,y_z\}$  : $$[a^6]\bigg(\frac{a}{1-a}\bigg)^6=1$$
By P.I.E : $$\binom{3}{1}\times 126 - \binom{3}{2}\times21 + \binom{3}{3} \times1 =316$$
Result : $$\frac{316}{426}=0,6839..$$
For generalization $$\frac{\sum_{i=1}^{N}\binom{N}{i}(-1)^{i+1}[a^{M}]\bigg(\frac{a}{1-a}\bigg)^{2i}\bigg(\frac{1}{1-a}\bigg)^{M-2i}}{[a^M]\bigg(\frac{1}{1-a}\bigg)^M}$$
A: $\newcommand{\mchoose}[2]{\left(\!\!\left(#1 \atop #2\right)\!\!\right)}\newcommand{\mchooset}[2]{\left(\!\left(#1 \atop #2\right)\!\right)}$The general problem, for an $m$-multisubset of $\{x_1,\dots,x_n,y_1,\dots,y_n\}$, can be handled using the principle of inclusion exclusion. I will first find the number of outcomes in the complementary event that there is no $i$ such that $x_i$ and $y_i$ both appear.
We start with all $\mchooset{2n}{m}$ multisets, and then subtract away the ones which contain both $x_i$ and $y_i$, for each $i\in \{1,\dots,n\}$. The result is
$$
\mchoose{2n}m-n\mchoose{2n}{m-2}
$$
However, we have doubly subtracted multisets which contain two pairs, $\{x_i,y_i\}$ and $\{x_j,y_j\}$, for $1\le i<j\le n$. For each of the $\binom n2$ ways to choose $\{i,j\}$, we need to add back in the $\mchooset{2n}{m-4}$ multisets which contain all four of $\{x_i,y_i,x_j,y_j\}$. So far, we are at
$$
\mchoose{2n}m-n\mchoose{2n}{m-2}+\binom{n}2\mchoose{2n}{m-4}
$$
This pattern of alternately adding and subtracting continues. The final result is
$$
\text{# multisets without $x_i,y_i$ for any $i$}=\sum_{i=0}^n(-1)^i\binom{n}i\mchoose{2n}{m-2i}
$$
To convert this to a probability, divide by $\mchooset{2n}m$, and remember to subtract from $1$ since we were dealing with the complementary event.
