The number of three-digit numbers with distinct digits that can be formed using the digits 1, 2, 3, 5, 8, and 9 is (options are: 24, 60, 120, 210, 336) the probability that both the first digit and the last digit of the three-digit number are even numbers is (options are: 1/6, 1/15, 2/15, 1/30, 1/42)

A) The number of three-digit numbers with distinct digits that can be formed using the digits 1, 2, 3, 5, 8, and 9 is; Option C: 120

B) The probability that both the first digit and the last digit of the three-digit number are even numbers is; Option B; 1/15

A) When we form such three-digit numbers with distinct digits using the digits 1, 2, 3, 5, 8, and 9 with all six digits different, and the order of digits doesn't matter, it means that we have six digits to permutate for 3 digit numbers by the formula;

⁶P₃ = 6!/(6 - 3)!

⇒ 6!/3!

⇒ 120 digits

B) Among all the six given numbers, there are only 2 which are even and they are 2 and 8.

If we choose 2  for the hundreds place, it means the units digit can only be 8 and any one of the other  4 numbers  can be used in the tens place. This means four choices.

Similarly with, eight in the hundreds place, we will have only 4 choices.

Thus, total number of choices = 8 possibilities.

Thus; probability that both the first digit and the last digit of the three-digit number are even numbers = 8/120 = 1/15

Read more about Permutation at; link

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

note the following:

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for

b and c

step-by-step explanation: