I don’t understand how to solve this with the tables.
1 answer:
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If A and B are equal:
Matrix A must be a diagonal matrix: FALSE.
We only know that A and B are equal, so they can both be non-diagonal matrices. Here's a counterexample:
Both matrices must be square: FALSE.
We only know that A and B are equal, so they can both be non-square matrices. The previous counterexample still works
Both matrices must be the same size: TRUE
If A and B are equal, they are literally the same matrix. So, in particular, they also share the size.
For any value of i, j; aij = bij: TRUE
Assuming that there was a small typo in the question, this is also true: two matrices are equal if the correspondent entries are the same.
Answer:
We can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.
Step-by-step explanation:
We have to find how many passwords with 4 different digits can we write with the numbers 1, 2, 3, 4, 5, and 6.
Firstly, it must be known here that to calculate the above situation we have to use Permutation and not combination because here the order of the numbers in a password matter.
Since we are given six numbers (1, 2, 3, 4, 5, and 6) and have to make 4 different digits passwords.
- Now, for first digit of the password, we have 6 possibilities (numbers from 1 to 6).
- Similarly, for second digit of the password, we have 5 possibilities (because one number from 1 to 6 has been used above and it can't be repeated).
- Similarly, for the third digit of the password, we have 4 possibilities (because two numbers from 1 to 6 have been used above and they can't be repeated).
- Similarly, for the fourth digit of the password, we have 3 possibilities (because three numbers from 1 to 6 have been used above and they can't be repeated).
So, the number of passwords with 4 different digits we can write = = 360 possibilities.
Hence, we can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.