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:1/15Step-by-step 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 .