MCQEasyJEE 2023Applications of P&C

JEE Mathematics 2023 Question with Solution

The number of square matrices of order 55 with entries from the set {0,1}\{0, 1\}, such that the sum of all the elements in each row is 11 and the sum of all the elements in each column is also 11, is:

  • A

    225225

  • B

    120120

  • C

    150150

  • D

    125125

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A square matrix of order 55 has entries from {0,1}\{0,1\}. Each row sum is 11 and each column sum is also 11.

Find: The number of such matrices.

Such a matrix must have exactly one entry equal to 11 in every row and exactly one entry equal to 11 in every column. Therefore, it is a permutation matrix.

Choose the position of 11 in the first row in 55 ways. Then for the second row, only 44 columns remain available. Similarly, for the third, fourth and fifth rows, the numbers of choices are 3,2,13,2,1 respectively.

Number of matrices=5×4×3×2×1\text{Number of matrices} = 5 \times 4 \times 3 \times 2 \times 1=5!=120= 5! = 120

Therefore, the number of such matrices is 120120. The correct option is B.

Illustrative figure for arranging exactly one entry equal to 1 in each row and each column of a 5 by 5 matrix.

Permutation Matrix Interpretation

Given: The matrix is of order 55, entries are only 00 or 11, and every row sum and column sum equals 11.

Find: How many such matrices exist.

A matrix satisfying these conditions has one and only one 11 in each row, and because each column sum is also 11, no two rows can place their 11 in the same column.

So the matrix is completely determined by assigning to each row a distinct column containing the entry 11. This is exactly a permutation of the 55 columns.

The number of such assignments is the number of permutations of 55 objects:

5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120

Hence, there are 120120 such matrices, and the correct option is B.

Common mistakes

  • Mistake: Counting only the choices for each row as 555^5. Why it is wrong: this ignores the condition that each column sum must also be 11. What to do instead: after choosing a column for one row, that column cannot be used again.

  • Mistake: Thinking the matrix only needs at least one 11 in each row and column. Why it is wrong: the sum in each row and each column is exactly 11, not greater than or equal to 11. What to do instead: enforce exactly one 11 per row and exactly one 11 per column.

  • Mistake: Not recognizing these as permutation matrices. Why it is wrong: the given row and column conditions are precisely the defining property of permutation matrices with entries in {0,1}\{0,1\}. What to do instead: convert the counting problem directly into counting permutations of 55 columns.

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