I need help with proving of 4 equalities from Combinatorics. Actually, they are quite obvious, but my teacher insists on the complete proof, "step-by-step" proof, but I have some troubles with it...
Given a set of $n$ elements $S=\{o_1, \ldots, o_n\}$ and $2$ partitions: $X = \{X_1, \ldots, X_k\}$ a partition of S into $k$ subsets, and $Y = \{Y_1, \ldots, Y_l\}$ - a partition of S into $l$ subsets.
We define the following:
$A_{11}$ - the number of pairs of elements in $S$ that are in the same subset in $X$ and in the same subset in $Y$
$A_{00}$ - the number of pairs of elements in $S$ that are in different subsets in $X$ and in different subsets in $Y$
$A_{10}$ - the number of pairs of elements in $S$ that are in the same subset in $X$ and in different subsets in $Y$
$A_{01}$ - the number of pairs of elements in $S$ that are in different subsets in $X$ and in the same subset in $Y$
Also we introduce the notations:
$N_{i,j} = |X_{i} \cap Y_{j}|$ - the number of elements into the intersection $X_{i}$ and $Y_{j}$
$A = {n \choose 2} = \frac{n(n-1)}{2}$ - the total number of pairs for $S$
$Q_{X} = \sum_{i}{X_{i} \choose 2}$
$Q_{Y} = \sum_{j}{Y_{j} \choose 2}$
It is needed to prove, that:
$A_{11} = \sum_{i,j}{N_{i,j} \choose 2}$
$A_{00} = A - Q_{X} - Q_{Y} + A_{11}$
$Q_{X} = A_{11} + A_{10}$
$Q_{Y} = A_{11} + A_{01}$
Write, please, the proof with all the details!
Sorry for my English.
Thanks in advance!