$n$ people are choosing numbers $X_1, \ldots , X_n$ randomly (uniformly) and independently, from: $\{1, 2, \ldots , \frac {n}{logn} \} $.
We'll notate $Y$ for the number of triplets $i<j<k$ such that $X_i=X_j=X_k$
We're required to find an asymptotic notation for $E[Y]$ and $Var(Y)$
Regarding the variance, I havent approached that yet.
For the expected value, I've defined a Bernoulli RV $X_i$ such that $X_i=1$ with probablity $P=\frac {logn}{n}$
Yet I'm not sure how to proceed from here, I was thinking of defining a Negative Binomial RV with $r=3$, though I dont know how to move on from that.
Any intuition/direction could help!
Hint: Let $m$ be the total number of triplets (you should be able to compute $m$, in terms of $n$).
Imagine we enumerate the $m$ triplets, in some predetermined order, from $1$ to $m$. Let $Z_t$ be the relevant indicator variable: $Z_t=1$ if the triplet verifies the condition $X_i=X_j=X_k$
Then $Y = \sum_{t=1}^m Z_t$
Then, to compute $E[Y]$ it's enough to compute $E[Z_t]$. which is easy.
The variance, is a little more complex.
Update: the number for triplets is $m=\binom{n}{3}$ (combinations) . Because the condition $i<j<k$ implies that the elements must be distinct and that, all all possible permutations we restrict to one (equivalently: we dont' take into account the order of the three items).
Now $E[Z_t]=P(X_i=X_j=X_k) = r \frac{1}{r^3}=1/r^2$ where $r=n/\log n$
Hence $$E[Y]= \frac{m}{r^2} = \binom{n}{3}\frac{\log ^2 n}{n^2}=\frac{n (n-1)(n-2)}{6}\frac{\log ^2 n}{n^2} \to \frac{n \log^2 n}{6}$$
Update 2: As for the variance. We want to compute $E[Y^2]$. But
$$E[Y^2]= \sum_{i=1}^m \sum_{j=1}^m E[Z_i Z_j]$$
where $E[Z_i Z_j]=P(Z_i=1,Z_j=1)$ Now, group those $m^2$ terms, according to how many elements the joined triplets comprise (from three to six). Compute $P(Z_i=1,Z_j=1)$ for each group and sum.