Intern
Joined: 26 Jan 2010
Status:casado
Posts: 44
Given Kudos: 1
Location: chile
Concentration: Educación
WE 1: <!-- m --><a class="postlink" href=""></a><!-- m -->
WE 2: <!-- m --><a class="postlink" href=""></a><!-- m -->
WE 3: <!-- m --><a class="postlink" href=""></a><!-- m -->
Re: A company plans to assign identification numbers to its empl
[#permalink]
19 Feb 2022, 13:10
Since each identification number consists of four numbers, then we have a row of four places to fill with digits.
How does it tell us that, THAT THESE FOUR DIGITS MUST BE DIFFERENT and that the first digit cannot be zero.
Applying the above:
For the first digit 9 candidates (the digits except zero).
For the second digit, we would have 10 possibilities (all digits), but they tell us that the digits must be different, since in the first digit we already occupy a digit and the second digit cannot be equal to the first, so for the second digit actually we have 9 possibilities.
For the third digit we would have 10 possibilities, but we already have one digit for the first place, and another digit for the second place, and since we are required to have different digits, for the third digit we have 10 -2 = 8, 8 possibilities .
For the fourth digit, following a similar reasoning to the previous ones, we had 7 possibilities.
Namely:
9 9 8 7
Okay. And what do I do with each of the numbers 9 9 8 7?
What would you do, multiply them or add them?
Stop for a moment and give an answer.
Very good!
We must multiply those numbers.
Why?
Because for all the candidates that are in one place, there are all the other candidates that are in the other places.
Then
9x9x8x7 = ?
Tips: To estimate a value of multiplication. When multiplying only pay attention to the unit.
This is 9x9 = 1, 1x8=8, 8x7=6
It is the unit of the number when multiplying 9x9x8x7 is 6, we see the alternatives and the only number ending in 6 is alternative B.
Answer B