Question

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

Answer 1

1/15

Explanation:

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 ⁶ P ₃ = 6 × 5 × 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 8 /120 = 1 /15

.