Three(3) digit numbers can be formed from the digits 0,1,3,5,7 and 8. How many numbers can be formed without repeats. (A) if the 3 digits numbers must be divisible by 5. (b) if the 3 digits number must be a multiple of 2
Your School SUBJECT is MATH.
consider this template for your 3 digit number
__________
| ? | ? | ? |
***********
A) the last digit can only be a 5 or 0, so two choices
the front can only be the 1,3,7, or 8, or 4 choices (can't be zero)
leaving 4 choices for the middle, now I could use the zero
__________
| 4 | 4 | 2 |
***********
number of such 3-digit numbers = 4x4x2 = 32
B) using the same kind of reasoning:
a multiple of 2, has to end in an even number, thus only one way, namely the 8
etc.
4x4x1 = 16
re-thinking part A
there are 2 cases,
1. the number ends in a 0
then there are 5 ways to start the front, then 4 ways to have the middle digit
number of ways = 5x4x1 = 20
2. then number ends in a 5
then there are 4 ways to start, no zero at the start
but then again 4 ways to have the middle, the zero can now be used
number of ways = 4x4x1 = 16
total is 20+16 = 36
To determine the number of three-digit numbers that can be formed without repeats using the digits 0, 1, 3, 5, 7, and 8, we can apply the principles of counting.
(a) If the three-digit number must be divisible by 5:
To be divisible by 5, the last digit of the three-digit number must be either 0 or 5. Let's consider both cases separately:
Case 1: The last digit is 0.
In this case, there are five options for the first digit (1, 3, 5, 7, or 8), and four options for the second digit since we cannot repeat any of the digits already chosen. Therefore, there are 5 × 4 = 20 possibilities.
Case 2: The last digit is 5.
In this case, we have the same five options for the first digit, and we only have three options for the second digit (as we cannot repeat any of the digits already chosen). Hence, there are 5 × 3 = 15 possibilities.
To get the total number of three-digit numbers that are divisible by 5, we sum the possibilities from both cases: 20 + 15 = 35.
Therefore, there are 35 three-digit numbers that can be formed without repeats and are divisible by 5.
(b) If the three-digit number must be a multiple of 2:
For a number to be a multiple of 2, its last digit must be an even number (0, 2, 4, 6, or 8). Let's consider the cases separately:
Case 1: The last digit is 0 or 8.
In each of these cases, we have four options for the first digit (1, 3, 5, or 7) and three options for the second digit (excluding the one already chosen). Thus, there are 2 × 4 × 3 = 24 possibilities for each case.
Case 2: The last digit is 2, 4, or 6.
In each of these cases, we also have four options for the first digit and three options for the second digit. Therefore, there are 3 × 4 × 3 = 36 possibilities for each case.
To find the total number of three-digit numbers that are multiples of 2, we sum the possibilities from all cases: 24 + 24 + 36 + 36 + 36 = 156.
Hence, there are 156 three-digit numbers that can be formed without repeats and are multiples of 2.