# How many numbers made out of 5 distinct digits contain 4 and are even

This is an exercise from a book on combinatorics and I can't seem to wrap my head around it: How many numbers are there, made out of 5 distinct digits, which contain '4' and are even. The answer according to the book is 7686.

I distinguished some cases (not sure if I distinguished in a 'smart' way)

• the last digit is 4. In that case, there are 8 possibilities for the first digit (not 0, not 4). The second digit has 8 possibilities (not the first, not 4), the third has 7, the fourth has 6. The total amount is $$8\cdot 8\cdot 7\cdot 6 = 2688$$.
• the last digit isn't 4. Therefore, the last digit must be 0,2,6 or 8. We must pick 3 digits from the remaining 8 digits. This can be done in $$8 \cdot 7 = 56$$ ways. These 3 digits along with '4' have to be distributed over the first 4 places, giving a total of $$56 \cdot 4 \cdot 4! = 5376$$ ways. However, these include some invalid numbers: those starting with $$0$$. These have to be substracted. The first digit is fixed, the last digit can be $$2,6,8$$, so 3 possibilities. We have to pick 2 digits from the remaining 7 digits. This can be done in $$7 \cdot 6/2 = 21$$ ways. We then need to distribute these two digits and '4' over 3 spaces, so $$3!$$ possibilities, giving a total of $$3 \cdot 21 \cdot 3! = 378$$ invalid numbers.

The total therefore equals $$2688 + 5376 - 378 = 7686$$ ways.

This seems like a brute force solution. Anyone who can 'smoothen' this, for example by making a smarter choice of distinguished cases?

I would break the problem into three main cases instead of two. There are $$5$$ positions for the $$4$$: the first position, the middle three, or the last position.

• If $$4$$ is in the first position, then the last position has $$4$$ possibilities to keep the number even ($$0$$, $$2$$, $$6$$, or $$8$$). In this case, you get the number of possibilities is: $$1\cdot 4\cdot 8\cdot 7\cdot 6=1344.$$ Here, the product order is "first, last, second, third fourth."

• If $$4$$ is in the last position, then the count above works. In this case, you get the number of possibilities is: $$1\cdot 8\cdot 8\cdot 7\cdot 6=2688.$$ Here, the product order is "last, first, second, third, fourth."

• If $$4$$ is in one of the middle positions, then the last position has $$4$$ possibilities. There are two possibilities here, either $$0$$ is the last position or it isn't.

• If $$0$$ is in the last position, then the number of possibilities is: $$3\cdot 1\cdot 8\cdot 7\cdot 6=1008.$$ Here, the product order is "position of $$4$$, last, first, two remaining positions."

• If $$0$$ is not in the last position, then the number of possibilities is: $$3\cdot 3\cdot 7\cdot 7\cdot 6=2646$$ Here, the product order is "position of $$4$$, last, first, two remaining positions."

Adding all of these up gives $$7686$$, as desired.

• I like this answer more than mine, it is much clearer. Thank you! – Student Nov 5 '18 at 17:47