The transpositions in the row stabilizer number
$$
\sum_i\binom{\lambda_i}{2}
=\sum_{(i,j)\in\lambda}(j-1),
$$
because a cell in column $j$ has $j-1$ cells before it in its row. Similarly, the transpositions in the column stabilizer number
$$
\sum_j\binom{\lambda'_j}{2}
=\sum_{(i,j)\in\lambda}(i-1).
$$
Their difference is
$$
\sum_{(i,j)\in\lambda}(j-i)
=\sum_{(i,j)\in\lambda}c_{i,j}(\lambda),
$$
the sum of the <Young-diagram cell contents>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 1 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-160/1/b/ii/solution