Let $S = \{A \in M_5(\Bbb{R}): a_{ij} = 0 \text{ or } 1 \text{ for all } i,j \text{ and } \\ \text{ ith row sum of $A$ = ith column sum of $A$ = $1$} \forall i \}$.

Then the number of elements in $S$ is

1. $5^2$,

2. $5^5$,

3. $5!$,

4. 55.

Then the number of elements in $S$ is

1. $5^2$,

2. $5^5$,

3. $5!$,

4. 55.

**Solution**: We will explicitly calculate the cardinality of $S$. We note that every matrix in $S$ is 0,1 matrix and has the row sum and column sum 1. Hence each row and each column has exactly one 1. Now, observe that any such matrix can be obtained from the $5 \times 5$ identity matrix by permuting the rows. This defines a bijection between the sets $S$ and $S_5$ of permutations on $\{1,2,3,4,5\}$. Such matrices are called permutation matrices and have cardinality $5!$.**Share to your groups:**

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