How Many Six-Digit Numbers Have Three Even and Three Odd Digits?

[Vineeth T]

Homework Statement

How many different six-digit numbers are there whose three digits are even and three digits are odd?

Homework Equations

No equations are required.We only need the principles of counting.

The Attempt at a Solution

I tried to split into 4 cases:
case I:there is no zero in the six digit number.such numbers =125*64*20
case II:there is one zero in the number.such numbers =125*16*5*10
case III:there are two zeroes in the number.such numbers =125*4*10*4
case IV:there are three zeroes in the number.such numbers =125*10
So the final answer will be 281250.
But the correct answer is 179550.
Can anyone show me where I have over counted ?

 

[HallsofIvy]

I did it a completely different way but get your answer. I break it into two cases:
1) The first digit is odd. In that case, there are 5 possible digits (1, 3, 5, 7, 9) for the first digit. There are then five possible digits for the other 5 digits as well. Also there are \begin{pmatrix}5 \\ 3\end{pmatrix}= 10 ways to place the three even digits in those 5 digits. There are 10(5^6)= 156250 ways in this case.

2) The first digit is even. In this case there are 4 possible digits (2, 4, 6, 8) since the first digit cannot be 0. There are again five possible digits for the other 5 digits and 10 ways to place the three odd digits in those 5 digts. There are 10(5^5)4= 125000 ways in this case.

Together there are 156250+ 125000= 281250 ways to do this.

 

[rcgldr]

grinding this out with a simple program confirms the total is 281250:

int main(int argc, char **argv)
{
int i, j, k, even, odd, total;

    total = 0;
    for(i = 100000; i < 1000000; i++){
        j = i;
        even = odd = 0;
        for(k = 0; k < 6; k++){ 
            if(j&1)
                odd += 1;
            else
                even += 1;
            j /= 10;
        }
        if((odd == 3) && (even == 3))
            total += 1;
    }
    printf("%d\n", total);
    return(0);
}

 

[Vineeth T]

Then is it confirm that the answer is 281250.
But the answer is given as 179550.(source:Problems In Mathematics With Hints And Solutions by V Govrov)
Thanks for everyone.

 

[rcgldr]

Also there are \begin{pmatrix}5 \\ 3\end{pmatrix}= 10 ways to place the three even digits in those 5 digits.

I think of this as the number of permutations for a multiset of 2 odds and 3 evens which is
\frac{5!}{2! \ 3!} = 10
{OOEEE, OEOEE, OEEOE, OEEEO, EOOEE, EOEOE, EOEEO, EEOOE, EEOEO, EEEOO}

Then is it confirm that the answer is 281250.

Assming that there isn't some additional condition that wasn't mentioned in the problem statement. If you understand the program example, does it match the problem statement?

You could also consider the number of cases allowing leading zeros - the number of case with leading zeros which is:

\frac{6!}{3! \ 3!} 5^6 \ - \ \frac{5!}{2! \ 3!} 5^5 = 312500 - 31250 = 281250