How many Young tableaux of size 6 are there? I have come up with the number 76, by counting the number of every possible Young tableau of weight 6, namely
{6}, {5,1}, {4,2}, {4,1,1}, {3,3}, {3,2,1}, {3,1,1,1}, {2,2,2}, {2,2,1,1}, {2,1,1,1,1}, {1,1,1,1,1,1}
, where every column corresponds with a row of the given Young tableau. My calculations are as follows (obtained by dividing 6! by all the correspondng hook lengths):
$\frac{6!}{2*3*4*5*6}+\frac{6!}{2*3*4*6}+\frac{6!}{2*2*4*5}+\frac{6!}{2*3*2*6}+\frac{6!}{2*2*3*3*4}+\frac{6!}{3*3*5}+\frac{6!}{2*3*2*6}+\frac{6!}{2*2*3*3*4}+\frac{6!}{2*2*4*5}+\frac{6!}{2*3*4*6}+\frac{6!}{2*3*4*5*6}=$
=1+5+9+10+5+16+10+5+9+5+1=76
Is this correct? And also, is there a simpler way to arrive at an answer? Since my calculation seems very cumbersome, and it's not always easy to count up all the different tableaux.
Recall that the Robinson-Schensted correspondence establishes a bijection between permutations in $S_n$ and pairs of Young tableaux of size $n$. This bijection has the property that if $\sigma \in S_n$ corresponds to the pair $(P, Q)$ then $\sigma^{-1}$ corresponds to $(Q, P)$. It follows that pairs $(P, P)$ of two identical Young tableaux correspond bijectively to permutations satisfying $\sigma = \sigma^{-1}$, which means that the number of Young tableaux of size $n$ is the number of involutions in $S_n$. This sequence is A000085 on OEIS which confirms that these two sequences are the same and also confirms your count of $76$; it begins
$$1, 1, 2, 4, 10, 26, \color{green}{76}, 232, 764, 2620, \dots$$
Involutions can be counted by considering their cycle decompositions; the result is that the number $I_n$ of involutions has exponential generating function
$$\sum_{n \ge 0} I_n \frac{x^n}{n!} = \exp \left( x + \frac{x^2}{2} \right)$$
which is more or less equivalent to the identity
$$I_n = \sum_{k=0}^{\left\lfloor \frac{n}{2} \right\rfloor} \frac{n!}{2^k k! (n-2k)!}$$
where the $k^{th}$ term of the RHS counts the number of involutions with $k$ two-cycles.