Question 828940: A 5 digit number divisible by 3 is to formed using numbers 0,1,2,3,4,5 without repetition .the total number of ways this can be done
Answer by Edwin McCravy(20054) (Show Source):
You can put this solution on YOUR website!
We use the fact that a number is divisible by 3 if and
only if the sum of its digits are divisible by 3.
Notice that 1+2+3+4+5 = 15, which is divisible by 3
The only other way we can have a sum of 5 digits
divisible by 3 is to replace the 3 by the 0 making
the sum 3 less:
1+2+0+4+5 = 12, which is divisible by 3
No other choice of 5 digits can have a sum divisible
by 3, because there is no other way to make the
sum 12 or 15, and we certainly can't have a sum of
9 or 18.
So the number of 5-digit numbers that can be formed
from the digits {1,2,3,4,5} is
choose the first digit 5 ways.
Choose the second digit 4 ways.
Choose the third digit 3 ways.
Choose the fourth digit 2 ways.
Choose the fifth digit 1 way.
5×4×3×2×1 = 5! = 120
And the number of 5-digit numbers that can be formed
from the digits {1,2,0,4,5} is figured this way.
We choose the most restrictive digit first, which is the
first digit. It can be chosen 4 ways, so
choose the first digit 4 ways. (It CANNOT be 0)
Choose the second digit 4 ways. (It CAN be 0)
Choose the third digit 3 ways.
Choose the fourth digit 2 ways.
Choose the fifth digit 1 way.
4×4×3×2×1 = 4×4! = 96
Answer: 120+96 = 216
Edwin
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