# probability

• Sep 9th 2013, 04:34 AM
Trefoil2727
probability
Find how many distinct number greater than 5000 and divisible by 3 can be formed from digits 3,4,5,6,and 0. Each digit being used at most once in any number.
• Sep 9th 2013, 05:39 PM
Soroban
Re: probability
Hello, Trefoil2727!

Quote:

Find how many distinct numbers greater than 5000 and divisible by 3 can be formed
from digits 3,4,5,6,and 0, each digit being used at most once in any number.

A number is divisible by 3 if its sum of digits is divisible by 3.

Four-digit numbers

The number begins with 5: .$\displaystyle 5\,\_\,\_\,\_$
The other 3 digits can be: .$\displaystyle \{0,3,4\},\:\{0,4,6\},\:\{3,4,6\}$
. . Each has $\displaystyle 3!$ permutations.
Hence, there are: $\displaystyle 3\cdot3! \,=\,18$ four-digit numbers that begin with 5.

The number begins with 6: .$\displaystyle 6\,\_\,\_\,\_$
The other 3 digits can be: .$\displaystyle \{0,4,5\},\:\{3,4,5\}$
. . Each has $\displaystyle 3!$ permutations.
Hence, there are: $\displaystyle 2\cdot3! \,=\,12$ four-digit numbers that begin with 6.

There are: $\displaystyle 18 + 12 \,=\,{\color{blue}30}$ four-digit numbers.

Five-digit numbers

The digits $\displaystyle \{0,3,4,5,6\}$ add up to 18, a multiple of 3.
Hence, any five-digit number will be divisible by 3.

There are $\displaystyle 4$ choices for the first digit.
The other 4 digits can be permuted in $\displaystyle 4!$ ways.
Hence, there are: $\displaystyle 4\cdot4! \,=\,{\color{blue}96}$ five-digit numbers.

Therefore, there are : $\displaystyle 30 + 96 \,=\,{\color{blue}126}$ such numbers.