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
3 answers
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
__________
| ? | ? | ? |
***********
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
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
2 answers