这些总结了本书中使用的符号 2 {\displaystyle 2} 3 {\displaystyle 3}
i 1 2 ϕ 1 ( i ) 1 2 ϕ 2 ( i ) 2 1 i 1 2 3 ϕ 1 ( i ) 1 2 3 ϕ 2 ( i ) 1 3 2 ϕ 3 ( i ) 2 1 3 ϕ 4 ( i ) 2 3 1 ϕ 5 ( i ) 3 1 2 ϕ 6 ( i ) 3 2 1 {\displaystyle {\begin{array}{c|cc}i&1&2\\\hline \phi _{1}(i)&1&2\\\phi _{2}(i)&2&1\end{array}}\qquad {\begin{array}{c|ccc}i&1&2&3\\\hline \phi _{1}(i)&1&2&3\\\phi _{2}(i)&1&3&2\\\phi _{3}(i)&2&1&3\\\phi _{4}(i)&2&3&1\\\phi _{5}(i)&3&1&2\\\phi _{6}(i)&3&2&1\end{array}}}
建议所有读者完成此练习。 建议所有读者完成此练习。 建议所有读者完成此练习。
问题 3
使用排列展开公式推导 3 × 3 {\displaystyle 3\!\times \!3}
解答
根据 例 3.6 给出以下内容。
| t 1 , 1 t 1 , 2 t 1 , 3 t 2 , 1 t 2 , 2 t 2 , 3 t 3 , 1 t 3 , 2 t 3 , 3 | = t 1 , 1 t 2 , 2 t 3 , 3 | P ϕ 1 | + t 1 , 1 t 2 , 3 t 3 , 2 | P ϕ 2 | + t 1 , 2 t 2 , 1 t 3 , 3 | P ϕ 3 | + t 1 , 2 t 2 , 3 t 3 , 1 | P ϕ 4 | + t 1 , 3 t 2 , 1 t 3 , 2 | P ϕ 5 | + t 1 , 3 t 2 , 2 t 3 , 1 | P ϕ 6 | = t 1 , 1 t 2 , 2 t 3 , 3 ( + 1 ) + t 1 , 1 t 2 , 3 t 3 , 2 ( − 1 ) + t 1 , 2 t 2 , 1 t 3 , 3 ( − 1 ) + t 1 , 2 t 2 , 3 t 3 , 1 ( + 1 ) + t 1 , 3 t 2 , 1 t 3 , 2 ( + 1 ) + t 1 , 3 t 2 , 2 t 3 , 1 ( − 1 ) {\displaystyle {\begin{array}{rl}{\begin{vmatrix}t_{1,1}&t_{1,2}&t_{1,3}\\t_{2,1}&t_{2,2}&t_{2,3}\\t_{3,1}&t_{3,2}&t_{3,3}\end{vmatrix}}&={\begin{aligned}&t_{1,1}t_{2,2}t_{3,3}\left|P_{\phi _{1}}\right|+t_{1,1}t_{2,3}t_{3,2}\left|P_{\phi _{2}}\right|\\&\quad +t_{1,2}t_{2,1}t_{3,3}\left|P_{\phi _{3}}\right|+t_{1,2}t_{2,3}t_{3,1}\left|P_{\phi _{4}}\right|\\&\quad +t_{1,3}t_{2,1}t_{3,2}\left|P_{\phi _{5}}\right|+t_{1,3}t_{2,2}t_{3,1}\left|P_{\phi _{6}}\right|\end{aligned}}\\&={\begin{aligned}&t_{1,1}t_{2,2}t_{3,3}(+1)+t_{1,1}t_{2,3}t_{3,2}(-1)\\&\quad +t_{1,2}t_{2,1}t_{3,3}(-1)+t_{1,2}t_{2,3}t_{3,1}(+1)\\&\quad +t_{1,3}t_{2,1}t_{3,2}(+1)+t_{1,3}t_{2,2}t_{3,1}(-1)\end{aligned}}\end{array}}}
问题 4
列出所有 4 {\displaystyle 4}
解答
这是所有 ϕ ( 1 ) = 1 {\displaystyle \phi (1)=1}
ϕ 1 = ⟨ 1 , 2 , 3 , 4 ⟩ ϕ 2 = ⟨ 1 , 2 , 4 , 3 ⟩ ϕ 3 = ⟨ 1 , 3 , 2 , 4 ⟩ ϕ 4 = ⟨ 1 , 3 , 4 , 2 ⟩ ϕ 5 = ⟨ 1 , 4 , 2 , 3 ⟩ ϕ 6 = ⟨ 1 , 4 , 3 , 2 ⟩ {\displaystyle {\begin{array}{rl}&\phi _{1}=\langle 1,2,3,4\rangle \quad \phi _{2}=\langle 1,2,4,3\rangle \quad \phi _{3}=\langle 1,3,2,4\rangle \\&\quad \phi _{4}=\langle 1,3,4,2\rangle \quad \phi _{5}=\langle 1,4,2,3\rangle \quad \phi _{6}=\langle 1,4,3,2\rangle \end{array}}} ϕ ( 1 ) = 1 {\displaystyle \phi (1)=1}
ϕ 7 = ⟨ 2 , 1 , 3 , 4 ⟩ ϕ 8 = ⟨ 2 , 1 , 4 , 3 ⟩ ϕ 9 = ⟨ 2 , 3 , 1 , 4 ⟩ ϕ 10 = ⟨ 2 , 3 , 4 , 1 ⟩ ϕ 11 = ⟨ 2 , 4 , 1 , 3 ⟩ ϕ 12 = ⟨ 2 , 4 , 3 , 1 ⟩ {\displaystyle {\begin{array}{rl}&\phi _{7}=\langle 2,1,3,4\rangle \quad \phi _{8}=\langle 2,1,4,3\rangle \quad \phi _{9}=\langle 2,3,1,4\rangle \\&\quad \phi _{10}=\langle 2,3,4,1\rangle \quad \phi _{11}=\langle 2,4,1,3\rangle \quad \phi _{12}=\langle 2,4,3,1\rangle \end{array}}} ϕ ( 1 ) = 3 {\displaystyle \phi (1)=3}
ϕ 13 = ⟨ 3 , 1 , 2 , 4 ⟩ ϕ 14 = ⟨ 3 , 1 , 4 , 2 ⟩ ϕ 15 = ⟨ 3 , 2 , 1 , 4 ⟩ ϕ 16 = ⟨ 3 , 2 , 4 , 1 ⟩ ϕ 17 = ⟨ 3 , 4 , 1 , 2 ⟩ ϕ 18 = ⟨ 3 , 4 , 2 , 1 ⟩ {\displaystyle {\begin{array}{rl}&\phi _{13}=\langle 3,1,2,4\rangle \quad \phi _{14}=\langle 3,1,4,2\rangle \quad \phi _{15}=\langle 3,2,1,4\rangle \\&\quad \phi _{16}=\langle 3,2,4,1\rangle \quad \phi _{17}=\langle 3,4,1,2\rangle \quad \phi _{18}=\langle 3,4,2,1\rangle \end{array}}} ϕ ( 1 ) = 4 {\displaystyle \phi (1)=4}
ϕ 19 = ⟨ 4 , 1 , 2 , 3 ⟩ ϕ 20 = ⟨ 4 , 1 , 3 , 2 ⟩ ϕ 21 = ⟨ 4 , 2 , 1 , 3 ⟩ ϕ 22 = ⟨ 4 , 2 , 3 , 1 ⟩ ϕ 23 = ⟨ 4 , 3 , 1 , 2 ⟩ ϕ 24 = ⟨ 4 , 3 , 2 , 1 ⟩ {\displaystyle {\begin{array}{rl}&\phi _{19}=\langle 4,1,2,3\rangle \quad \phi _{20}=\langle 4,1,3,2\rangle \quad \phi _{21}=\langle 4,2,1,3\rangle \\&\quad \phi _{22}=\langle 4,2,3,1\rangle \quad \phi _{23}=\langle 4,3,1,2\rangle \quad \phi _{24}=\langle 4,3,2,1\rangle \end{array}}}
问题 6
证明 f {\displaystyle f} v → , w → ∈ V {\displaystyle {\vec {v}},{\vec {w}}\in V} k 1 , k 2 ∈ R {\displaystyle k_{1},k_{2}\in \mathbb {R} }
f ( ρ → 1 , … , k 1 v → 1 + k 2 v → 2 , … , ρ → n ) = k 1 f ( ρ → 1 , … , v → 1 , … , ρ → n ) + k 2 f ( ρ → 1 , … , v → 2 , … , ρ → n ) {\displaystyle f({\vec {\rho }}_{1},\dots ,k_{1}{\vec {v}}_{1}+k_{2}{\vec {v}}_{2},\dots ,{\vec {\rho }}_{n})=k_{1}f({\vec {\rho }}_{1},\dots ,{\vec {v}}_{1},\dots ,{\vec {\rho }}_{n})+k_{2}f({\vec {\rho }}_{1},\dots ,{\vec {v}}_{2},\dots ,{\vec {\rho }}_{n})}
解答
对于“如果”部分，第一个条件来自取 k 1 = k 2 = 1 {\displaystyle k_{1}=k_{2}=1} k 2 = 0 {\displaystyle k_{2}=0}
“当且仅当”部分也很常规。从 f ( ρ → 1 , … , k 1 v → 1 + k 2 v → 2 , … , ρ → n ) {\displaystyle f({\vec {\rho }}_{1},\dots ,k_{1}{\vec {v}}_{1}+k_{2}{\vec {v}}_{2},\dots ,{\vec {\rho }}_{n})} = f ( ρ → 1 , … , k 1 v → 1 , … , ρ → n ) + f ( ρ → 1 , … , k 2 v → 2 , … , ρ → n ) {\displaystyle =f({\vec {\rho }}_{1},\dots ,k_{1}{\vec {v}}_{1},\dots ,{\vec {\rho }}_{n})+f({\vec {\rho }}_{1},\dots ,k_{2}{\vec {v}}_{2},\dots ,{\vec {\rho }}_{n})}
问题 7
求此矩阵的排列展开式中唯一的非零项。
| 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 | {\displaystyle {\begin{vmatrix}0&1&0&0\\1&0&1&0\\0&1&0&1\\0&0&1&0\end{vmatrix}}}
解答
为了在排列展开中得到一个非零项，我们必须使用 1 , 2 {\displaystyle 1,2} 4 , 3 {\displaystyle 4,3} 2 , 1 {\displaystyle 2,1} 3 , 4 {\displaystyle 3,4} ⟨ 2 , 1 , 4 , 3 ⟩ {\displaystyle \langle 2,1,4,3\rangle } + 1 {\displaystyle +1}
( 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 ) {\displaystyle {\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}}} ρ 1 ↔ ρ 2 {\displaystyle \rho _{1}\leftrightarrow \rho _{2}} ρ 3 ↔ ρ 4 {\displaystyle \rho _{3}\leftrightarrow \rho _{4}}
问题 8
如果我们将定义中的性质 (4) 改为 | I | = 2 {\displaystyle \left|I\right|=2}
解答
它们都会翻倍。
问题 9
验证 [线性代数/排列展开#cor:ColSwapChgSign](/wiki/Linear_Algebra/The_Permutation_Expansion#cor:ColSwapChgSign) 中的第二和第三条语句。
解答
对于第二条语句，给定一个矩阵，将其转置，交换行，再转置回来。结果是交换了列，行列式变化了一个因子 − 1 {\displaystyle -1}
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问题 11
判断真假：一个矩阵，其元素只有 0 或 1，它的行列式等于 0、1 或 -1。([Strang 1980](#CITEREFStrang1980))
解答
假。
| 1 − 1 1 1 | = 2 {\displaystyle {\begin{vmatrix}1&-1\\1&1\end{vmatrix}}=2}
问题 12
证明一个 5 × 5 {\displaystyle 5\!\times \!5} 120 {\displaystyle 120} 如果 1 , 2 {\displaystyle 1,2}
解答
第一行元素的列索引有五种选择。然后，第二行元素的列索引有四种选择（第一行使用的列索引不能在此使用）。继续下去，我们得到 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120 {\displaystyle 5\cdot 4\cdot 3\cdot 2\cdot 1=120} 一旦我们选择第一行的第二列，我们就可以用 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 {\displaystyle 4\cdot 3\cdot 2\cdot 1=24}
问题 13
有多少个 n {\displaystyle n}
解答
n ⋅ ( n − 1 ) ⋯ 2 ⋅ 1 = n ! {\displaystyle n\cdot (n-1)\cdots 2\cdot 1=n!}
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问题 15
为了确保一个 4 × 4 {\displaystyle 4\!\times \!4}
解答
证明三个零的任何放置都不足以使行列式为零是常规的。四个零足以使行列式为零；将它们全部放在同一行或同一列。
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问题 16
如果我们有 n {\displaystyle n} ( x 1 , y 1 ) , ( x 2 , y 2 ) , … , ( x n , y n ) {\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),\dots \,,(x_{n},y_{n})} p ( x ) = a n − 1 x n − 1 + a n − 2 x n − 2 + ⋯ + a 1 x + a 0 {\displaystyle p(x)=a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots +a_{1}x+a_{0}} n {\displaystyle n} n {\displaystyle n}
| 1 1 … 1 x 1 x 2 … x n x 1 2 x 2 2 … x n 2 ⋮ x 1 n − 1 x 2 n − 1 … x n n − 1 | {\displaystyle {\begin{vmatrix}1&1&\ldots &1\\x_{1}&x_{2}&\ldots &x_{n}\\{x_{1}}^{2}&{x_{2}}^{2}&\ldots &{x_{n}}^{2}\\&\vdots \\{x_{1}}^{n-1}&{x_{2}}^{n-1}&\ldots &{x_{n}}^{n-1}\end{vmatrix}}} 等于所有索引 i , j ∈ { 1 , … , n } {\displaystyle i,j\in \{1,\dots ,n\}} i < j {\displaystyle i<j} x j − x i {\displaystyle x_{j}-x_{i}} x i {\displaystyle x_{i}}
解答
n = 3 {\displaystyle n=3} − x 1 ρ 2 + ρ 3 {\displaystyle -x_{1}\rho _{2}+\rho _{3}} − x 1 ρ 1 + ρ 2 {\displaystyle -x_{1}\rho _{1}+\rho _{2}}
| 1 1 1 x 1 x 2 x 3 x 1 2 x 2 2 x 3 2 | = | 1 1 1 x 1 x 2 x 3 0 ( − x 1 + x 2 ) x 2 ( − x 1 + x 3 ) x 3 | = | 1 1 1 0 − x 1 + x 2 − x 1 + x 3 0 ( − x 1 + x 2 ) x 2 ( − x 1 + x 3 ) x 3 | {\displaystyle {\begin{vmatrix}1&1&1\\x_{1}&x_{2}&x_{3}\\x_{1}^{2}&x_{2}^{2}&x_{3}^{2}\end{vmatrix}}={\begin{vmatrix}1&1&1\\x_{1}&x_{2}&x_{3}\\0&(-x_{1}+x_{2})x_{2}&(-x_{1}+x_{3})x_{3}\end{vmatrix}}={\begin{vmatrix}1&1&1\\0&-x_{1}+x_{2}&-x_{1}+x_{3}\\0&(-x_{1}+x_{2})x_{2}&(-x_{1}+x_{3})x_{3}\end{vmatrix}}} 然后， x 2 ρ 2 + ρ 3 {\displaystyle x_{2}\rho _{2}+\rho _{3}}
= | 1 1 1 0 − x 1 + x 2 − x 1 + x 3 0 0 ( − x 1 + x 3 ) ( − x 2 + x 3 ) | = ( x 2 − x 1 ) ( x 3 − x 1 ) ( x 3 − x 2 ) {\displaystyle ={\begin{vmatrix}1&1&1\\0&-x_{1}+x_{2}&-x_{1}+x_{3}\\0&0&(-x_{1}+x_{3})(-x_{2}+x_{3})\end{vmatrix}}=(x_{2}-x_{1})(x_{3}-x_{1})(x_{3}-x_{2})}
问题 17
矩阵可以分成块 ，例如
( 1 2 0 3 4 0 0 0 − 2 ) {\displaystyle \left({\begin{array}{cc|c}1&2&0\\3&4&0\\\hline 0&0&-2\end{array}}\right)} 它显示了四个块，左上角和右下角是 2 × 2 {\displaystyle 2\!\times \!2} 1 × 1 {\displaystyle 1\!\times \!1}
T = ( J Z 2 Z 1 K ) {\displaystyle T=\left({\begin{array}{c|c}J&Z_{2}\\\hline Z_{1}&K\end{array}}\right)} 其中 J {\displaystyle J} K {\displaystyle K} Z 1 {\displaystyle Z_{1}} Z 2 {\displaystyle Z_{2}} | T | = | J | ⋅ | K | {\displaystyle \left|T\right|=\left|J\right|\cdot \left|K\right|}
解答
设 T {\displaystyle T} n × n {\displaystyle n\!\times \!n} J {\displaystyle J} p × p {\displaystyle p\!\times \!p} K {\displaystyle K} q × q {\displaystyle q\!\times \!q}
∑ permutations ϕ {\displaystyle \sum _{{\text{permutations }}\phi }} | T | = ∑ permutations ϕ t 1 , ϕ ( 1 ) t 2 , ϕ ( 2 ) … t n , ϕ ( n ) | P ϕ | {\displaystyle \left|T\right|=\sum _{{\text{permutations }}\phi }t_{1,\phi (1)}t_{2,\phi (2)}\dots t_{n,\phi (n)}\left|P_{\phi }\right|} 因为 T {\displaystyle T} ϕ {\displaystyle \phi } p + 1 , … , n {\displaystyle p+1,\dots ,n} p {\displaystyle p} ϕ ( 1 ) , … , ϕ ( p ) {\displaystyle \phi (1),\dots ,\phi (p)} ϕ {\displaystyle \phi } 0 {\displaystyle 0} ϕ ( 1 ) = n {\displaystyle \phi (1)=n} t 1 , ϕ ( 1 ) t 2 , ϕ ( 2 ) … t n , ϕ ( n ) {\displaystyle t_{1,\phi (1)}t_{2,\phi (2)}\dots t_{n,\phi (n)}} 0 {\displaystyle 0} 1 , … , p {\displaystyle 1,\dots ,p} p + 1 , … , p + q {\displaystyle p+1,\dots ,p+q} | J | ⋅ | K | {\displaystyle \left|J\right|\cdot \left|K\right|}
[ ∑ perms ϕ 1 of 1 , … , p t 1 , ϕ 1 ( 1 ) ⋯ t p , ϕ 1 ( p ) | P ϕ 1 | ] ⋅ [ ∑ perms ϕ 2 of p + 1 , … , p + q t p + 1 , ϕ 2 ( p + 1 ) ⋯ t p + q , ϕ 2 ( p + q ) | P ϕ 2 | ] {\displaystyle {\bigg [}\sum _{\begin{array}{c}\\[-19pt]\scriptstyle {\text{perms }}\phi _{1}\\[-5pt]\scriptstyle {\text{of }}1,\dots ,p\end{array}}t_{1,\phi _{1}(1)}\cdots t_{p,\phi _{1}(p)}\left|P_{\phi _{1}}\right|{\bigg ]}\cdot {\bigg [}\sum _{\begin{array}{c}\\[-19pt]\scriptstyle {\text{perms }}\phi _{2}\\[-5pt]\scriptstyle {\text{of }}p+1,\dots ,p+q\end{array}}t_{p+1,\phi _{2}(p+1)}\cdots t_{p+q,\phi _{2}(p+q)}\left|P_{\phi _{2}}\right|{\bigg ]}}
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问题 18
证明对于任何 n × n {\displaystyle n\!\times \!n} T {\displaystyle T} n {\displaystyle n} r {\displaystyle r} T − r I {\displaystyle T-rI}
解答
n = 3 {\displaystyle n=3}
| T − r I | = | t 1 , 1 − x t 1 , 2 t 1 , 3 t 2 , 1 t 2 , 2 − x t 2 , 3 t 3 , 1 t 3 , 2 t 3 , 3 − x | {\displaystyle \left|T-rI\right|={\begin{vmatrix}t_{1,1}-x&t_{1,2}&t_{1,3}\\t_{2,1}&t_{2,2}-x&t_{2,3}\\t_{3,1}&t_{3,2}&t_{3,3}-x\end{vmatrix}}} 排列展开式中的每一项都包含从矩阵中取出的三个因子（例如， ( t 1 , 1 − x ) ( t 2 , 2 − x ) ( t 3 , 3 − x ) {\displaystyle (t_{1,1}-x)(t_{2,2}-x)(t_{3,3}-x)} ( t 1 , 1 − x ) ( t 2 , 3 ) ( t 3 , 2 ) {\displaystyle (t_{1,1}-x)(t_{2,3})(t_{3,2})} x {\displaystyle x} 3 {\displaystyle 3}
一般来说，排列展开式表明行列式可以写成一项之和，每一项都有 n {\displaystyle n} n {\displaystyle n} n {\displaystyle n} n {\displaystyle n}
问题 20
证明
| x − 2 x − 3 x − 4 x + 1 x − 1 x − 3 x − 4 x − 7 x − 10 | = 0. {\displaystyle {\begin{vmatrix}x-2&x-3&x-4\\x+1&x-1&x-3\\x-4&x-7&x-10\end{vmatrix}}=0.} (Silverman & Trigg 1963 )
解答
这是引用的来源中给出的答案。
当任何一列的元素从其他两列的每个元素中减去时，导出的行列式的两列中的元素成比例，因此行列式消失。也就是说，
| 2 1 x − 4 4 2 x − 3 6 3 x − 10 | = | 1 x − 3 − 1 2 x − 1 − 2 3 x − 7 − 3 | = | x − 2 − 1 − 2 x + 1 − 2 − 4 x − 4 − 3 − 6 | = 0. {\displaystyle {\begin{vmatrix}2&1&x-4\\4&2&x-3\\6&3&x-10\end{vmatrix}}={\begin{vmatrix}1&x-3&-1\\2&x-1&-2\\3&x-7&-3\end{vmatrix}}={\begin{vmatrix}x-2&-1&-2\\x+1&-2&-4\\x-4&-3&-6\end{vmatrix}}=0.}
? 问题 22
证明帕斯卡三角形左上角 n 2 {\displaystyle n^{2}}
1 1 1 1 . . 1 2 3 . . 1 3 . . 1 . . . . {\displaystyle {\begin{array}{cccccc}1&1&1&1&.&.\\1&2&3&.&.\\1&3&.&.&&\\1&.&.&&&\\.\\.\end{array}}} 的值为一。（Rupp & Aude 1931 ）
解答
这是引文来源中给出的答案。 用 D n {\displaystyle D_{n}} a i , j {\displaystyle a_{i,j}} i {\displaystyle i} j {\displaystyle j}
a i , j = a i , j − 1 + a i − 1 , j , a 1 , j = a i , 1 = 1. {\displaystyle a_{i,j}=a_{i,j-1}+a_{i-1,j},\qquad a_{1,j}=a_{i,1}=1.} 从 D n {\displaystyle D_{n}} n − 1 {\displaystyle n-1} a i , j {\displaystyle a_{i,j}} a i , j − 1 {\displaystyle a_{i,j-1}} a 1 , 1 = 1 {\displaystyle a_{1,1}=1} a i , j − 1 {\displaystyle a_{i,j-1}} a i − 1 , j − 1 {\displaystyle a_{i-1,j-1}} a i , j {\displaystyle a_{i,j}} a i − 1 , j − 1 {\displaystyle a_{i-1,j-1}} D n = D n + i {\displaystyle D_{n}=D_{n+i}}
D n = D n − 1 = D n − 2 = ⋯ = D 2 = 1. {\displaystyle D_{n}=D_{n-1}=D_{n-2}=\dots =D_{2}=1.}
Silverman, D. L. (proposer); Trigg, C. W. (solver) (1963), "Quickie 237", Mathematics Magazine , American Mathematical Society, 36 (1) Strang, Gilbert (1980), Linear Algebra and its Applications (2nd ed.), Hartcourt Brace Javanovich Trigg, C. W. (proposer) (1963), "Quickie 307", Mathematics Magazine , American Mathematical Society, 36 (1): 77 Trigg, C. W. (proposer); Walker, R. J. (solver) (1949), "Elementary Problem 813", American Mathematical Monthly , American Mathematical Society, 56 (1) Rupp, C. A. (proposer); Aude, H. T. R. (solver) (1931), "Problem 3468", American Mathematical Monthly , American Mathematical Society, 37 (6): 355 
![{\displaystyle {\bigg [}\sum _{\begin{array}{c}\\[-19pt]\scriptstyle {\text{perms }}\phi _{1}\\[-5pt]\scriptstyle {\text{of }}1,\dots ,p\end{array}}t_{1,\phi _{1}(1)}\cdots t_{p,\phi _{1}(p)}\left|P_{\phi _{1}}\right|{\bigg ]}\cdot {\bigg [}\sum _{\begin{array}{c}\\[-19pt]\scriptstyle {\text{perms }}\phi _{2}\\[-5pt]\scriptstyle {\text{of }}p+1,\dots ,p+q\end{array}}t_{p+1,\phi _{2}(p+1)}\cdots t_{p+q,\phi _{2}(p+q)}\left|P_{\phi _{2}}\right|{\bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b0d9b75cdd834e9d7eb1b2431ba8461ef3e5819)
