Mathematics, 28.08.2019 12:00 Richelleypinelley

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)

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Answer from: oasmith2005

There are six digits to choose from, but you're only taking three at a time, so the number of such numbers is

${}_6P_3=\dfrac{6!}{3!}=6\times5\times4=120$

The first and last digits can only be even if the number takes the one of the forms "2 _ 8" or "8 _ 2". The middle number can be any of the remaining four, so there is a total of eight such numbers.

This means the probability of getting a number beginning and ending with an even digits is $\dfrac8{120}=\dfrac1{15}$ .

Answer from: nmooesy2229

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

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