Suppose we have a pack of $52$ cards. Each card has a number from $1$ to $13$ and there are $4$ identical cards for each number. We are searching for the number of splittings, consisting of $7$ cards, where a player has at least one triad of the same card, but not $4$ identical cards.
I did consider specific cases to find the number of splittings. So :
a) Two triads of the same number (e.g. $1,1,1,2,2,2,5$ ) $$ n_1 = 11{13 \choose 2} $$
b) One triad and the rest are all different (e.g. $1,1,1,2,4,6,7$ ) $$ n_2 = 13{12 \choose 4} $$
c) One triad, one pair and the rest are different (e.g. $1,1,1,3,3,6,7$ ) $$ n_3 = 13 \cdot 12 {11 \choose 2} $$
d) One triad and two pairs (e.g. $1,1,1,3,3,6,6$ ) $$ n_4 = 13 {12 \choose 2} $$
So, the answer should be $ N = n_1 + n_2 + n_3 + n_4 = 16731$.
Is the above correct ?
Is there another (perhaps faster) way to solve this ?
Thank you in advance
Each admissible selection of $7$ cards produces a partition of the number $7$ into parts of size $\leq3$, among them at least one part of size $3$. There are $4$ such partitions, namely $$3+3+1,\quad 3+2+2, \quad 3+2+1+1,\quad 3+1+1+1+1\ .$$ You have gone through all of them and computed for each the number of possible choices of the entries. At the end you obtained a total of $16\,731$. If you solve this problem by hand I don't see a faster method.
In order to check your result I considered the power series $$p(x):=(1+x+x^2+x^3)^{13}+O[x]^8,\qquad q(x):=(1+x+x^2)^{13}+O[x]^8$$ which are the generating functions of the number of choices containing at most triads, resp., at most pairs of equals from $[1\,..\,13]$. Then I let Mathematica compute the coefficient of $x^7$ in the series $p-q$. The result was $16\,731$ as well.