How many even numbers greater than 50 000 can be formed from specified digits without repeat?

Question

The digits are 3,4,5,6,7,0

My working is as follows:

I realize that you would need to start with either 5,6 or 7. From there you have 5 digits to re-arrange, but the permutation would have to end in an even number.

Starting with 5 you would end with either 4, 6 or 0; Thus you would have 4 digits left to permute and 3 even numbers to end with so: (4!)(3)

Starting with 6 you would end with 4 or 0; Thus (4!)(2)

Starting with 7 would be the same as with 5.

In the end my answer was (4!)(3) + (4!)(2) + (4!)(3) = 192. The answer given in my text book is 504.

Any help is appreciated. Thanks.

Answer 1

Add up the following:

• Amount of 5-digit numbers that start with $5$ and end with $0$ is $\binom{4}{3}\cdot3!$

• Amount of 5-digit numbers that start with $5$ and end with $4$ is $\binom{4}{3}\cdot3!$

• Amount of 5-digit numbers that start with $5$ and end with $6$ is $\binom{4}{3}\cdot3!$

• Amount of 5-digit numbers that start with $6$ and end with $0$ is $\binom{4}{3}\cdot3!$

• Amount of 5-digit numbers that start with $6$ and end with $4$ is $\binom{4}{3}\cdot3!$

• Amount of 5-digit numbers that start with $7$ and end with $0$ is $\binom{4}{3}\cdot3!$

• Amount of 5-digit numbers that start with $7$ and end with $4$ is $\binom{4}{3}\cdot3!$

• Amount of 5-digit numbers that start with $7$ and end with $6$ is $\binom{4}{3}\cdot3!$

• Amount of 6-digit numbers that start with anything but $0$ and end with $0$ is $5!$

• Amount of 6-digit numbers that start with anything but $0$ and end with $4$ is $5!-4!$

• Amount of 6-digit numbers that start with anything but $0$ and end with $6$ is $5!-4!$

So the overall amount is $24+24+24+24+24+24+24+24+120+96+96=504$.

Answer 2

The number can have 5 or 6 digits and can not begin with 0

1. the numbers with 5 digits being even; 1*4*3*2*2 =48
2. the numbers with 5 digits being odd; 3*4*3*2*2=144
3. the numbers with 6 digits being even; 2*4*3*2*1*2=96
4. the numbers with 6 digits being odd; 3*4*3*2*1*3=216 Therefore there are 48+144+96+216=504 possibilities