# How many three-digit numbers with distinct digits can be formed using the digits 1, 2, 3, 5, 8, and 9? What is the probability that both the first digit and the last digit of the three digit number are even numbers?

Jun 27, 2017

We can make $120$ three digit distinct numbers. The probability, that both the first digit and the last digit of the three digit number are even numbers, is $\frac{1}{15}$.

#### Explanation:

Here I am assuming that digits can be used once only.

When we form such three-digit numbers with distinct digits using the digits $1 , 2 , 3 , 5 , 8$ and $9$ (in all six digits all different), observe that the order of digits does matter. For example, if we make a number using digits $1 , 2 ,$ and $3$, we can have $123 , 132 , 231 , 213 , 312$ or $321$.

Hence we have to find number of $3$-digit numbers that can be made from these six digits using permutation and answer is ${P}_{3}^{6} = 6 \times 5 \times 4 = 120$.

How haw many of them will have first digit as even, we have two choices $2$ and $8$. Once we have chosen $2$ for hundreds place, we can have only $8$ in units place and any one of remaining $4$ can be used in tens place. Hence four choices, with $2$ in hundreds place and another four choices when we have $8$ in hundreds place (and $2$ in units place) i.e. total $8$ possibilities.

Hence, the probability, that both the first digit and the last digit of the three digit number are even numbers, is $\frac{8}{120} = \frac{1}{15}$.