An example of application of the Littlewood–Richardson rule [closed]

Question

I am computing the Littlewood–Richardson coefficients (https://en.wikipedia.org/wiki/Littlewood%E2%80%93Richardson_rule) of the product $s_{[2,2]}s_{[1,1]}$ both by hand and a software tool (https://www.jgibson.id.au/articles/symfunc/). When computing by hand to me, it appears that $s_{[2,2,2]}$ should be in the product because the skew tableaux of the shape [2,2,2]/[2,2] is just a row with two blocks and placing 1,2 in that order ([1][2]) to the table gives us a Littlewood–Richardson tableaux with content [1,1]. But with software tool [2,2,2] does not appear in the product. What goes wrong with my calculation.

Answer 1

the skew tableaux of the shape [2,2,2]/[2,2] is just a row with two blocks and placing 1,2 in that order ([1][2]) to the table gives us a Littlewood–Richardson tableaux with content [1,1]

This is not correct. This isn't a Littlewood-Richardson tableau because the rightmost entry is a 2. In any Littlewood-Richardson tableau, the rightmost entry of the top row has to be a 1 for the following reason.

According to (one) definition of Littlewood-Richardson tableau, when you read the elements of the tableau from right to left starting from the top row, then the second row, etc. then if you stop at any point, the number of occurrences of $i$ must be >= the number of occurrences of $j$ whenever $i < j$. In particular, if you stop at the first step, you must have a 1 because if it was some other value $i$ there would be 0 1's and 1 $i$ which contradicts the definition.