0

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.

Graeme Harris
  • 3
  • I think you forgot that you have 6 digits to use. The first number is indeed 5 digits starting with 5, ending with 6 (50346), but the highest one is 765430 (6 digits)... You need to compute the number of 5 digits numbers plus the number of 6 digits ones, that can start with 3 and 4... EDIT: If you HAVE to use the six numbers, the 0 can be placed first to do 5 digits numbers, but they can't finish with 0... – Martigan Jul 08 '14 at 15:10
  • Thank you, it seems so obvious on a second glance! I realized my mistake this morning after reviewing the question again but thank you for the answer. :) – Graeme Harris Jul 09 '14 at 19:59

2 Answers2

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 $24+24+24+24+24+24+24+24+120+96+96=504$.

barak manos
  • 43,109
0

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
Moses Nielsen
  • 1