Permutations

ian.f

Just a quick question on permutations and the like. It isn't too challenging but I am looking for a handier way of doing it.

There are 2 parts to the question. It's the 2nd part i want help with but obviously you will need to see the first to understand it.

1) How many 4 digit numbers can you make for the numbers 1 -> 9 if there is no repetition? 9x8x7x6 = 3024

2) How many of these numbers are in ascending order e.g 1234, 1358 etc.

I know this can be done by manually grinding out each example by saying how many of them start with one, how many start with 2 etc but I'm looking for a way that won't take all morning.... any ideas?

NB: It is obvious that none of the numbers can start with 7,8,9 but after that it just gets a bit tedious

aurthurg

Probably should post my solution

so the amount of ascending combinations starting with 1

8X7X6 = 336


starting with 2

7X6X5 = 210

each time you lose the first number and add the next lowest in the sequence (e.g the next line we lose the 7 and multiply by 5) then add them up, am sure there is matheamatical expression so this can be applied to any amount of no.s i'll try my best but i should be studyin

aurthurg

o.k i really hope i have nt mess this entire question up but am thinkin this describes the soln

for four digit no.s

sigma (n=9) to (n=0) of (n-1)(n-2)(n-3)

for five digit no.s

sigma (n=9) to (n=0) of (n-1)(n-2)(n-3)(n-4)

and so on

ian.f

Does multiplying them by that not put them in a random fashion (ie no guarantee of ascending)

i.e in the case of starting with one if you multiply by 8x7x6 are you not getting 336 random combinations which may or may not be ascending

ian.f

Cheers for the formula... I don't have the answer sheet on me now to check but seems to be good stuff.. Still don't understand how it guarantees that they are ascending but I'll take your word for it for the time being.

aurthurg

yes sorry i overlooked someting i think the answer is actually 1512

the formula only gives the number ascending begginning with 1 i'l try and work on it

aurthurg

Edit: In the complete disgust of not seeing the simplicity of this problem

Crumbs

For part 2 of the question, order matters. That's a clue to use combinations.

Edit: Removing my rubbish answer. Just read MathsManiac's post below.

MathsManiac

There seem to be a lot of very complicated ways of looking at this above!

The answer is simply 9C4 = 126, since every selection of four distinct digits from the nine available gives precisely one arrangement in ascending order.

ian.f

126 is the right answer according to the solution sheet. Nice one... The problem with things like this is giving the problem too much respect



Edit: Just read your solution properly there and I understand completely now. Fair play maths maniac