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


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


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?

  • 1
    $\begingroup$ Are those two a,b different questions or conditions $\endgroup$ – Archis Welankar Jan 10 '16 at 12:40
  • $\begingroup$ Conditions. I have fixed the question. Thanks. $\endgroup$ – A nobody Jan 10 '16 at 12:41


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.

  • $\begingroup$ 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? $\endgroup$ – A nobody Jan 10 '16 at 13:02
  • $\begingroup$ There is no rule I followed. You could call it mathematical maturity :). $\endgroup$ – drhab Jan 10 '16 at 14:06
  • $\begingroup$ No worries. Thanks. I suppose more practice then. $\endgroup$ – 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 $4\times8 \frac{8!}{4!}+\frac{9!}{4!} > 5\times 8\frac{8!}{4!}$

  • $\begingroup$ @drhab, oh yes! thanks. i will edit it. $\endgroup$ – 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


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