Hello
This is my first post.
The question I'm working on is, 'All the digits of a number are different, the first digit is not zero, and the sum of the digits is 36. There are N x 7! such numbers. What is the value of N?'
I think I have figured out that 1,9,2,8,3,7,6 gives me 7! Then there is 3,7,4,6,2,8,1,5, which gives me 8!. I can add and factor to get 9 x 7!. Now I need to include the zero. That gives me both 8 and 9 digit numbers but the zero can't be in the first place. I haven't studied permutations and am not sure how to proceed. I think I would do 8! and 9! respectively and subtract something but I can't get my head around that difference.
If someone could point me in the right direction it would be very much apprecaited.
Thank you.
Plato, thank you very much for taking the time to reply. It's very helpful. Yes that is what I am working on. I thought that perhaps I had not yet found all the combinations of different digits with a sum of 36, but the zero was bothering me more, thank you for the pointer. I will check I haven't missed any other combinations. May I ask, is the number of arrangements of n different digits, when one of the digits cannot be in one place, (n-1) (n-1)!?
Hello Plato
Thanks to you I was able to finish the problem.
I certainly had missed a few combinations and should have given it more thought before posting originally!
I got:
87654321
9876321
9875421
9865431
9765432
987651
987642
987543
8! + (4)(7!) + (3)(6!)
876543210
98763210
98754210
98654310
97654320
9876510
9876420
9875430
(8)8! + (28)(7!) + (18)(6!)
Arrangements:
(9)8! + (32)(7!) + (21)(6!)
7![(9)(8) +32 +3]
7!(107)
N = 107
Thanks again for your help.