How many even 3 digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?

Answer 1

It depends on whether you allow a leading zero or not.
If you are allowed a leading 0, its easy. The last digit can be 0,2,4, or 6 (4 possibilities), and after you've chosen that, theres 7 possibilities for the first digit and 6 possibilities for the second digit, which makes 4*7*6 = 168.
However, usually 3 digit numbers can't have leading zeroes. Now we have to split it up into two cases:
Case 1:
_ _ 0
There are 7 choices for the first digit, and 6 choices for the second - 42 possibilities.
Case 2:
It ends with a 2, 4, or 6.
There are now 3 choices for the last digit, 6 choices for the first digit (anything *but* 0), and 6 choices for the second digit - 3*6*6 = 108.
So the total is 150.

Note that we dealt with the 0 case specially, because it made a difference whether we had a 0 to work with for the first digit or not.

(Note - again, in response to Wal C below, you *have* to split it up into cases. He said 4*6*6; however, if you happened to choose the 0 for the last digit as part of that '4', there will be *7* possibilities for the first digit, not 6. So you can't do it all together like that.)

Answer 2

FIRST: How many 3 digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7 if no digit is repeated?

( 8 * 7 * 6 )
three digits, all digits must be different

SECOND: How many of those combinations are even?
8 digits, half even, half odd
So half of the time the third digit will be even and the number will be even.

(The combinations you lose when even digits ar 'used up' in the first two places are balanced by those 'extra' combinations when odd digits start the number.)

Rick

Answer 3

In total there are 8 different digits

you are making 3 digit numbers. I shall arbitrarily decide that numbers beginning with a zero are prohibited....(068 is not a valid choice).

[If you decide to make numbers with leading zeros valid, there are 8 ways of choosing the first place.]

There ar 7 ways of choosing the first digit. .........(0 is forbidden)

There are now 7 digits to choose from in the second place........(0 permitted)

There are now 6 ways of choosing the third digit..........(2 already taken)


The total number of ways you can make a 3 digit number is

7 x 7 x 6 = 294
Of these only half are even,

So there are 147 even 3 digit numbers

Answer 4

Now they must be even so they must end in 0, 2, 4 or 6

So if allowed to start with 0
No of possibilities = 4 * 7 * 6 (4 possibilities of last digit * 7 left for 1st * 6 left for 2nd)
= 168

If NOT allowed to start with 0
No of possibilities = 4 * 6 * 6 (4 possibilities of last digit * 6 left for 1st (no 0) * 6 left for 2nd)
= 144

Answer 5

Well, to make it even, what choices do you have for the last digit?

And once you pick that digit, how many choices for the other 2 digits, if no digits are repeated?

Multiply it all together to get the answer.

Answer 6

case 1: with 0 * * *
7 6 1=42
case2: with 2 * * *
6 6 1=36
case 3: with 4 * * *
6 6 1=36
case 4: with 6 * * *
6 6 1=36
total = 42+36*3=150 3ven dig without rep.