Given all four digits odd numbers.How many of them have exactly two even numbers?

The sample set is $ \{0,1,2,3,4,5,6,7,8,9\}$. To have a odd number, the last digit have to be odd, so there are five possible ways to the last digit. In order to have a four digit number, the first can't be zero. In a four digit number, if exactly two are even, the other two have to be odd( one of them is the last). So, I made a permutation with replacement regarding to the five odd numbers and two positions.The result was $5^2=25$.

In other hand, the total of four digit odd numbers is $9\cdot 10^2 \cdot 5$. Nine for the possible ways to the first digit, $10^2$ for the second and third digit and $5$ to the last.

At last I subtracted $25$ from $4500$.However the result don't match to the solutions.Can you help me? Thanks

  • 1
    $\begingroup$ Your solution starts correctly, but I don't see what you mean with "So, I made a permutation with replacement ... The result was 25". It should be clear that there are more than 25 numbers you should substract from all the 4-digit numbers that end with an odd digit to get what you want. $\endgroup$ – MichalisN Jan 16 '12 at 21:59

I think this is the simplest approach:

There are four place holders to fill.

The last digit [ones digit] should be odd and therefore, there are $5$ ways of filling this slot.

Now, the thousands digit can be odd or even.

In case, it is odd, you can fill this in $5$ ways, but forcing you to fill the other two digits with an even number which can be done in $5^2$ ways.

If it is even, since it cannot be $0$, it can be filled in $4$ ways. And in $\binom{2}{1} \cdot 5$ ways, one of the other digits will be filled with an even number and the left-over must be a odd digit and hence can be filled in $5$ ways.

So, the number of numbers satisfying your criteria is-$5^4+5\times4\times\binom{2}{1}\times5\times5=1625$

As you point in your comments, I have looked at disjoint cases: EEOO, EOEO,OEEO where E stands for even and O stands for odd.

Hope this helps.

  • 1
    $\begingroup$ I hunderstood. I have to study 3 diferent situations.Odd-Even-Even-Odd ; Even-Odd-Even-Odd ; Even-Even-Odd-Odd. And them apply to fundamental count rule, been attention to the first digit restrition.At the end I sum the 3 situations.The result was 1625 $\endgroup$ – João Jan 17 '12 at 1:00
  • $\begingroup$ @João I fixed it up. I had written things correctly and in the final step, I forgot to multiply by 5. Thank You. $\endgroup$ – user21436 Jan 17 '12 at 1:15

Here's another approach.

5 ways for the last odd digit & 5*3 ways to choose & place other odd digit.

Positions for even digits have got fixed, so 5^2 ways for even digits, giving 5*5*3*5^2 = 1875

from this, subtract #s starting with 0 [ 1/5 th such in 2 of 3 configurations = 2/15 ]

ans: 1875 *13/15 = 1625


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