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I posted a question on this, and the number of matrices came to be 36:

The number of $3\times3$ non singular matrices with four entries as 1 and all other entries as 0, is (JEE Main 2010)

However, I also came across this, which counts the same thing for three by three matrices with exactly four elements being an $a$, the answer there is 45.

I am trying to reconcile the two. What is different in the second question that makes the answer 45 (as compared to 36).

Starlight
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  • Can you clarify with an example where one is different from the other? Because if you have one element in each row and each column, that implies invertibility (at least for three by three matrices). – Starlight May 13 '24 at 02:22
  • For the second question, why can’t we have two rows that are (1,1,0) and the last row being all zeroes? Is that invertible? – Randall May 13 '24 at 02:26
  • The question says, "I am trying to count the number of n×n matrices, where every entry is a lowercase alphabet, such that every row and column contains at least one "a". Your example has the last row zero, which contradicts that. – Starlight May 13 '24 at 02:28
  • Ah, I see now. Good catch. – Randall May 13 '24 at 02:30

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Some $3$ by $3$ zero-one matrices have exactly four ones with at least one one in each row and column but are singular. For example, $$\begin{pmatrix} 0 &0 &1 \\ 0 &0 &1 \\ 1 &1 &0 \end{pmatrix}$$ So your two questions address different sets of matrices. There are $45$ $3$ by $3$ zero-one matrices having exactly four ones with at least one one in each row and column. Of these, $36$ are non-singular and $9$ are singular.

awkward
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