How Many Even Integers?

Before we split into cases, notice one thing: given any $m$ numbers of a set of $n$ integers ( $m \le n$ ), there is exactly one permutation per $m$ different numbers chosen that has all $m$ numbers in strictly increasing order. Thus, the number of ways we can arrange $m$ elements of a set of $n$ integers in stricly increasing order is equal to $\binom{n}{m}$ .

We'll break cases into the possibilities of the last [units] digit. Obviously, the last digit cannot be either $0$ or $2$ , so we have only $3$ cases:

Case 1: The last digit is $4$ . This obviously yields $1$ case.

Case 2: The last digit is $6$ . By our logic above, we can choose $3$ elements from the set $\{ 1, 2, 3, 4, 5 \}$ to be the first three digits and arrange them, which yields $\binom{5}{3} = 10$ cases.

Case 3: The last digit is $8$ . Again by our logic above, we can choose $3$ elements from the set $\{ 1, 2, 3, 4, 5, 6, 7 \}$ to be the first three digits and arrange them, which yields $\binom{7}{3} = 35$ cases.

Summing our cases, we find that there are $1 + 10 + 35 = \boxed{46}$ such numbers.

There are only three cases wherein we can assemble the four-digit number with it being an even number. It's only when the last digit is 4, 6 or 8. Since the digits must be strictly increasing, the last digit must be the greatest of the 4. Case 1: When the last digit is 4 There are only three natural numbers less than 4. We cannot put zero as a digit because the digits must be strictly increasing. So to assemble the number, there are 3C3=1 way. Case 2: When the last digit is 6 The same as the first. There are 5 natural numbers less than 6. To choose the other 3 digits, there are 5C3=10 ways. Case 3: When the last digit is 8 There are 7C3=35 ways. Summing up all the cases, we have 1+10+35=46 total number of four-digit numbers

The numbers can only have 4,6 or 8 at the end, because these are the only even single-digit numbers which have at least 3 positive integers(>0) less than them.

With 4 at the end, there's only one possible combination so that the numbers increase from left to right: 1234

With 6 at the end, you can choose three numbers out of (1,2,3,4,5) to fill for the preceding 3 digits. It would be tempting to use permutations of the 5 numbers, with 3 objects. But that would give the wrong answer, because taking permutations takes all the possible arrangements of the three selected numbers. But for the 3 selected numbers, we require only one arrangement - * the one where the numbers are arranged in ascending order *

So, what we do is we select any 3 combinations from the 5 integers, and arrange them in the desired way. So, all we need to do is use combinations. $5C_{3} =10$

With 8 at the end, we have to choose 3 numbers out of (1,2,3,4,5,6,7) which will be $7C_{3}=35$

The total sum is $1+10+35=46$

If the last digit is 0 or 2,number of required numbers=0.

If the last digit is 4,the digits 1,2,and 3 can be used as the other digits.Thus there will be one number like that.

If the last digit is 6,the digits 1,2,3,4 and 5 can be used as the other digits.Among them 3 can be chosen in 10 ways.

If the last digit is 8,the digits 1,2,3,4,5,6 and 7 can be used as the other digits.Among them 3 can be chosen in C(7,3) ways i.e,35 ways

Therefore total number of required 4-digit numbers=1+10+35= 46

Let the digits of the four digit number be a,b,c,and d If d=4, a,b,and c can only have 3C3 ways to operate If d=6 a,b, and c can only have 5C3 ways to operate if d=8 a,b, and c can only have 7C3 ways to operate Therefore, we add all our cases to get 46 total ways I hope this was a nice solution.=)