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)

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 ________. (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)

anonymous · Answer

I got 120 for the first part, now i need help on the 2nd part

zepdrix · Answer

Are we allowed to repeat digits on the second one?
The wording is sort of confusing :( Hmm
I can't figure out if it's connected to the first question or not.

anonymous · Answer

no repeats is how i got my answer for the first one

anonymous · Answer

if i did repeats, the answer wasnt one of the options

zepdrix · Answer

ya 120 seems right for the first one :)
I was thinking about the second hehe

anonymous · Answer

Yay :) and im assuming the 2nd part has no repeats?

zepdrix · Answer

Well we only have 3 even numbers to in our selection.
So 3! out of the 120 perhaps?

zepdrix · Answer

Hmm that's not an option >.<

anonymous · Answer

No :(

zepdrix · Answer

Oh oh, derp. my bad

anonymous · Answer

first and last digit of a 3 digit number have to be even

zepdrix · Answer

We can place any of the 6 numbers in the middle.
Then we have 2 numbers left which can go in the first and second slot in any order.
6 x 2! ya?

zepdrix · Answer

No no that didn't work either XD haha

anonymous · Answer

any 6 numbers can go in the middle, and theres only 2 even numbers

zepdrix · Answer

oh wow I was counting a third somehow! >.<
good catch

zepdrix · Answer

In order to get an even in both slots 1 and 3,
we CAN NOT use an even in the middle then, ya?

So 5 options for the middle, then 2! for the other two slots

anonymous · Answer

Yes

zepdrix · Answer

10/120 = 1/12,
not an option...
ok back to the drawing board :)

anonymous · Answer

theres 6 numbers, 2 are even, so only 4 can be in the middle?

zepdrix · Answer

yes good call :)
boy my brain is just not working tonight, sorry

anonymous · Answer

Lol its okay

zepdrix · Answer

That seems to match one of our options ya? :)

anonymous · Answer

i dont know the answer yet

zepdrix · Answer

oh ok sorry :) i'm jumping the gun

4 for the middle, then permutate the last 2,
2! x 4 = 8

probability is that many divided by the total, ya? 8 / 120

anonymous · Answer

so 1/15 :)

zepdrix · Answer

yay team \c:/
I think we did it

anonymous · Answer

yes thanks :)

zepdrix · Answer

For these types of problems, drawing out a diagram usually works for me.|dw:1451448629225:dw|