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This might be a classic question, but since I am new to representation theory of the symmetric group, I am asking it here.

Suppose that $n=k+l+r$ where $k\geq l\geq r\geq 0$. Let $G$ be the symmetric group $S_n$ and $H$ be its Young subgroup $S_k \times S_l \times S_1^r$. What can be said about the induced representation $\mbox{Ind}_H^G 1$? What are its components? Can we say something about their multiplicities?

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    $\begingroup$ This is called the permutation module associated to the Young subgroup. The decomposition into irreducibles is given by the Kostka numbers - they have a nice combinatorial description. You can read a nice exposition in Sagan's book "The symmetric group". $\endgroup$ – Noah White Aug 8 '17 at 17:45
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The answer is a special case of Young's rule. In my book, I give a very simple method for the slightly easier case where $r=0$. In that case we have: $$ \mathrm{Ind}_{S_k\times S_l}^{S_n} = \bigoplus_{s=0}^l V_{(n-s, s)}, $$ a multiplicity-free decomposition, where $V_{(n-s, s)}$ is an irreducible representation of dimension $\binom ns - \binom n{s-1}$.

In your more general case, the irreducible constituents will all be Specht modules of the form $V_{(s, t, u)}$, where $s\geq k$, $s+t\geq k+l$, and (of course) $s+t+u=k+l+r$. The multiplicities are given by Kostka numbers, which can be greater than one. For example, if $(k, l, r) = (3, 2, 1)$, then $V_{(5, 1)}$ occurs with multiplicity $2$.

The multiplicity of $V_{(s, t, u)}$ with $(s, t, u)=(5, 1, 0)$ can be calculated in sage as, for example:

sage: sage.libs.symmetrica.all.kostka_number([5, 1], [3, 2, 1])
2

More generally, replace [5, 1] with [s, t, u].

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