Perms and combs question (1 Viewer)

[bleakarcher]

3 numbers are chosen from 1,2,3,4,5,6,7,8,9.

i) How many ways that 3 numbers can be chosen in order of increasing magnitude?
ii) How many ways 3 numbers be picked such that their sum is 10?

Thanks guys. This question popped into my prelim exam today.

 

[RealiseNothing]

Hint for first one: triangular numbers.

 

[RealiseNothing]

I think (i) is:



Where denotes the k'th triangular number.

 

[bleakarcher]

I put down 9C3 for the first part. Fuck.

 

[Sindivyn]

I thought the first one was 9C3 tbqh

 

[Shadowdude]

While that might work - it'll probably be shot down in the HSC Exam.

Use a less airy-fairy method and if all else fails, list cases.

 

[bleakarcher]

I thought the first one was 9C3 tbqh

Yeah it makes sense. Pick any three and once you have done so arrange them in order of increasing magnitude which can only be done 1 way.

 

[Shadowdude]

Yeah it makes sense. Pick any three and once you have done so arrange them in order of increasing magnitude which can only be done 1 way.

How so?

You misunderstood the question. It's not about arranging, it's about whether you get them in ascending order.

So 1 2 3 is valid

But 2 3 1 - in that order - isn't.

 

[RealiseNothing]

Another way of doing it is this:

Consider you only have the number 1, 2, and 3. How many ways can you arrange these?

1 2 3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1

How many of these are increasing? Only 1. Hence for the numebrs 1, 2, and 3, the amount increasing is the total arrangements divided by 6.

If we extend this to a genral solution for any 3 numbers, we can have:

Lowest Middle Highest

Lowest Highest Middle

Middle Lowest Highest

Middle Highest Lowest

Highest Lowest Middle

Highest Middle Lowest

Once again, there is only one that increases from left to right. So the general solution is the amount of 3 letter codes you can make, divided by 6.

Hence for your question, it would be:

 

[RealiseNothing]

Just putting it out there:

9C3 is 84, which is still correct. I just did it in a way that makes it seem intuitive.

 

[deswa1]

How so?

You misunderstood the question. It's not about arranging, it's about whether you get them in ascending order.

So 1 2 3 is valid

But 2 3 1 - in that order - isn't.

Yeah but that's his point. You can choose a set of three numbers in 9C3 ways, so that will leave you with the set {1,2,3} and then you can arrange each given set in exactly ONE way were the integers are increasing -> therefore you can choose 9C3 ways of ascending numbers

 

[Shadowdude]

Yeah but that's his point. You can choose a set of three numbers in 9C3 ways, so that will leave you with the set {1,2,3} and then you can arrange each given set in exactly ONE way were the integers are increasing -> therefore you can choose 9C3 ways of ascending numbers

oh yeah that's right

mixed up my Cs and Ps


time to sleep

making silly maffs errors

 

[bleakarcher]

Yeah but that's his point. You can choose a set of three numbers in 9C3 ways, so that will leave you with the set {1,2,3} and then you can arrange each given set in exactly ONE way were the integers are increasing -> therefore you can choose 9C3 ways of ascending numbers

yes this.

 

[bleakarcher]

Second part I put down 9C2 which is definitely wrong. I thought about it more after the exam and got (7C2)-1. Not sure though.

 

[Sindivyn]

Second one you can completely discount 8 and 9. Tbqh, I would count afterwards (unless there's another way I can't see).

 

[deswa1]

For part ii) can numbers be repeated? I'm assuming not but want to make sure before I do it

 

[RealiseNothing]

Well you can only have {1,2,7} {1,3,6} {1,4,5} or {2,3,5}.

 

[bleakarcher]

For part ii) can numbers be repeated? I'm assuming not but want to make sure before I do it

They can't be repeated.