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What is the number of all possible matrices of order 2 $\times $ 3 with each entry 0, 1?

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Hint: Calculate the number of elements that can be there in the matrix of order 2 $\times $ 3. These elements can have only 0 or 1. Using the concept of permutations and combinations to find the number of all possible matrices of order 2 $\times $ 3 with each entry 0 or 1.

Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In matrices, if we have a matrix of order m $\times $ n, then the number of elements in this matrix are equal to mn . . . . . . . . . . . . . . . . (1)
In permutations and combination, by the concept of principle of counting, if there are n places and on each place, we can place m numbers, then, the number of ways in which we can have different numbers is equal to ${{m}^{n}}$. . . . . . . . . . . . . . (2)
In this question, we are given a matrix of order 2 $\times $ 3. Using formula (1), we can say that the number of elements in this matrix is equal to (2)(3) = 6.
Also, in the question, it is given that at every place, the matrix can have either 0 or 1. So, using formula (2), the number of possible matrices is equal to ${{2}^{6}}=64$.
Hence, the answer is 64.

Note: There is a possibility that one may commit a mistake while using the formula (2). It is possible that one may apply the formula ${{n}^{m}}$ instead of the formula ${{m}^{n}}$ which will give us an incorrect answer.
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