how many 4 digit even numbers without repitition are there?

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- Sep 13th 2012, 08:39 PM #1

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- Sep 13th 2012, 09:52 PM #2

- Sep 14th 2012, 01:53 AM #3

- Sep 14th 2012, 04:38 PM #4

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## Re: problem...

Hello, vandna281!

How many 4-digit even numbers without repetition are there?

I assume that a number may not begin with zero.

There are five even digits: 0, 2, 4, 6, 8.

Then there are two cases to consider.

[1] The number ends in zero: _ _ _ 0

. . .Then the remaining digits can be selected in 9·8·7 ways.

. . .Hence, there are 504 numbers that end in zero.

[2] The number does not end in zero: _ _ _ e

. . .There are 4 choices for "e".

. . .There are 8 choices for the first (leftmost) digit; any digit except 0 and "e".

. . .Then there are 8·7 choices for the remaining digits.

. . .Hence, there are: 4·8·8·7 = 1792 numbers that do not end in zero.

Therefore, there are: 504 + 1792 = 2296 four-digit even numbers

. . which agrees with Kalyanram's answer.

- Sep 14th 2012, 05:44 PM #5

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## Re: problem...

Four digits forces the lead digit to be in {1, 2, ..., 9}. Next two digits can be any in {0, 1, 2, ..., 9}. Last digit must be in {0, 2, 4, 6, 8) in order to be even.

Thus there are 9 x 10 x 10 x 5 = 4500 different such even four digit numbers.

If by distinct you mean all four digits are distinct, then:

Choosing the distinct first & last digits first, and then each of the digits in between.

The total number of even 2 digit numbers is 9x5 = 45 (pick the first digit, then the last digit). Repetitions occur at 22, 44, 66, 88.

Thus there are 41 even two digit numbers without digit repetition. Now split each of them into the first and last digit of our 4 digit numbers: xy -> x _ _ y.

We've 8 choices for the 3rd digit that will avoid digit replication. Then 7 choices for the 2nd (and final) digit that will avoid digit replication.

Thus the number of even four digits numbers where all four digits are distinct is 41 x 8 x 7 = 2296.