What Is The Number Of Even Numbers Greater Than 100 That Can Be Formed With 0 1 2 3?

Because we want even numbers, the 1s digit must be 0 or 2.
When the 1s digit is 0, we can freely choose the remaining digits. There are 6 permutations of 3 digits taken 2 at a time (to make a 3-digit number ending in 0), and 6 permutations of 3 digits taken 3 at a time (to make a 4-digit number ending in 0). Thus, there are 12 ways to make a 3- or 4-digit number ending in 0 from the given digits.

When the 1s digit is 2, we need to choose the 100s digit of the 3-digit numbers from the set {1, 3}. The 10s digit can be either of the remaining two digits, for a total of 2*2 = 4 ways to make a 3-digit number ending in 2.

The same restriction applies for the 4-digit numbers ending in 2: The 1000s digit must be from the set {1, 3}, and the remaining digits are free choice. Again, there are 2*2*1 = 4 ways to choose those.

We have a total of 12 numbers ending in 0, and 8 numbers ending in 2, for a total of 20 even numbers greater than 100 made from digits {0, 1, 2, 3}.

The entire list is
120, 130, 210, 230, 310, 320, 1230, 1320, 2130, 2310, 3120, 3210,
102, 132, 302, 312, 1032, 1302, 3012, 3102.
thanked the writer.
Hey dear!
The answer 20 is not correct because without repetition number
of even numbers is 10.
The numbers are 102, 132, 302, 312, 120, 130, 210, 230, 310,
and 320.
The numbers with repetitions are infinite if it is more than 100 but if it is more than hundred but within three digits, then the answer is 23.
The numbers are 102, 112, 122, 132, 202, 212, 222, 232, 302, 312, 322, 332, 110, 120, 130, 200, 210, 220, 230, 300, 310, 320 and 330.
thanked the writer.
Anonymous commented
Can you please help me with this by solving without writing the possibilities but by using permutation method? 