MATH SOLVE

2 months ago

Q:
# 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 .

Accepted Solution

A:

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
.