# Permutations- the number of six digit integers that are even

Determine the number of six digit integers in which no digit may be repeated and the integers are even.

I understand how to do this when we are repeating digits

$9*10^4*5$

When repetition is not allowed, the total number of integers is given by

$\frac{9*9*8*7*6*5}{2}$

Would it be correct to divide the above by $2$. I basically figured that because the total is even, dividing by $2$ will give me the total number of even integers. But, is there another way to get to the right answer using permutation formulas?

• Are those two a,b different questions or conditions – Archis Welankar Jan 10 '16 at 12:40
• Conditions. I have fixed the question. Thanks. – A nobody Jan 10 '16 at 12:41

Hint:

Determine $N_0,N_2,N_4,N_6,N_8$ where $N_i$ denotes the number of five digit integers in wich no digit may be repeated and digit $i$ is not one of the used digits.

Then you are looking for $N_0+N_2+N_4+N_6+N_8$.

Further on it is evident that $N_2=N_4=N_6=N_8$ so it is enough to find $N_0$ and $N_2$.

The idea is that every five digit integers in which no digit may be repeated and in which e.g. digit $2$ is not allowed ($N_2$ exist) will be made a six digit even integer by adding a $2$ on the right.

• Ok, thanks. So, $N_0= 9*8*7*6*5$ and $N_2=8*8*7*6*5$. I multiply $N_2 * 4$ and add $N_0$ to get $\frac{9!}{5!}+\frac{8!}{4!}*8$ to get $68880$. Wonderful. How do you know to solve it this way? Is it simply experience, or are there rules for approaching these problems that I don't know? – A nobody Jan 10 '16 at 13:02
• There is no rule I followed. You could call it mathematical maturity :). – drhab Jan 10 '16 at 14:06
• No worries. Thanks. I suppose more practice then. – A nobody Jan 10 '16 at 14:11

Another way

Since it is easier to compute odd numbers, compute

All numbers - Odd numbers = $9*9*8*7*6*5 - 8*8*7*6*5*5 = 68,880$

Dividing by 2 shouldn't be correct because intuitively the number of even numbers will be more than number of odd numbers. This is because in the even numbers the $0$ is "more free" (it can sit in the units place as well whereas in the odd numbers $0$ has only 4 places to sit in.

To rigourise this intuition lets count:

$Case 1$: $0$ is in the unit place. Then we have $\frac{9!}{4!}$ many numbers

$Case 2$: $0$ isn't in the unit place. Then we have 4 numbers for the units place and 8 numbers for the leftmost digit. So, we have $4\times8 \frac{8!}{4!}$.

So, answer is $4\times8 \frac{8!}{4!}+\frac{9!}{4!} > 5\times 8\frac{8!}{4!}$

• @drhab, oh yes! thanks. i will edit it. – Subham Jaiswal Jan 10 '16 at 15:14

Split the set of all integers into 3 disjoint subsets:

1) ending on 0, 2) no 0 in the integer, 3) 0 present in 4 slots other than the last

1) You have 9 values for 5 slots, $\binom{9}{5} \cdot 5!$

2) You have 4 values for the last slot and 8 for the other 5

3) yoy have 4 values for the ladt slot, 0 anywhere in 4 and 8 values for 4 slots

Now sum them up