An Arrangement Problem for Fun

[Debater Slayer]

Consider the numbers from 1 to 10. In how many ways is it possible to build a sequence of 5 numbers such that all the numbers are in ascending order?

Please make sure to post your answer behind a spoiler tag!

 

[ChristineM]

Consider the numbers from 1 to 10. In how many ways is it possible to build a sequence of 5 numbers such that all the numbers are in ascending order?

Please make sure to post your answer behind a spoiler tag!

This is fun?

 

[Debater Slayer]

This is fun?

Well, if not, it helps to pretend it is while solving.

 

[VoidCat]

Well, if not, it helps to pretend it is while solving.

*raises hand* what if you dont know how to go about the problem? Can i pretend to have fun while pretending to solve the problem?

 

[Debater Slayer]

*raises hand* what if you dont know how to go about the problem? Can i pretend to have fun while pretending to solve the problem?

Sure! There's the added bonus of not being bound by exam time or grades.

 

[Secret Chief]

This is fun?

My thought entirely. I've spent most of today in maths lessons and this is what I come home to.

 

[Secret Chief]

Sure! There's the added bonus of not being bound by exam time or grades.

Party animal.

 

[ChristineM]

Well, if not, it helps to pretend it is while solving.

I did try then started jumping numbers 12456 etc and lost the will to live

 

[vulcanlogician]

I did try then started jumping numbers 12456 etc and lost the will to live

Me too. I'm hoping that there is some trick or algorithm to do it (that OP knows but we have to figure out).

Otherwise, it seems a lil' tedious.

@Debater Slayer Is there a trick to doing it or not? If so, figuring that out DOES seem like fun.

 

[Debater Slayer]

Me too. I'm hoping that there is some trick or algorithm to do it (that OP knows but we have to figure out).

Otherwise, it seems a lil' tedious.

@Debater Slayer Is there a trick to doing it or not? If so, figuring that out DOES seem like fun.

Yes, there is a systematic method to solve this using a branch of mathematics known as combinatorics.

That's a hint! Don't tell anyone or they'll cheat the quizz.

 

[vulcanlogician]

That's a hint! Don't tell anyone or they'll cheat the quizz.

My lips are sealed.

 

[ImmortalFlame]

Possible combinations starting with 6: 1
Possible combinations starting with 5: 5
Possible combinations starting with 4: 10
Possible combinations starting with 3: 20

I'm sure someone could figure out a rule from here?

 

[beenherebeforeagain]

I did try then started jumping numbers 12456 etc and lost the will to live

It's a problem that Marvin would probably be very despondent about...

 

[It Aint Necessarily So]

Isn't this asking how many permutations there are of ten things taken five at a time?

Spoiler: suggested solution

P(n,r)=n! (n−r)! = 10! / 5! = 30240

Possible combinations starting with 6: 1
Possible combinations starting with 5: 5
Possible combinations starting with 4: 10
Possible combinations starting with 3: 20

I also tried doing this using brute force. Our result concur until starting with 3:

6 - 1 (6789T)
5 - 5 (56789, 5678T, 5679T, 5689T, 5789T)
4 - 10 (45678, 45679, 4567T, 45689, 4568T, 4569T, 45789, 4578T, 4579T, 4589T)
3 - 35 (34567, 34568, 34569, 3456T; 34578, 34579, 3457T, 34589, 3458T, 3459T; 34678, 34679, 3467T, 34689, 3468T, 3469T; 34789, 3478T, 3479T; 3489T; 35678, 35679, 3567T, 35689, 3568T; 3569T; 35789, 3578T, 3579T, 3589T, 36789, 3678T, 3679T, 3689T, 3789T)

 

[ImmortalFlame]

Isn't this asking how many permutations there are of ten things taken five at a time?

Spoiler: suggested solution

P(n,r)=n! (n−r)! = 10! / 5! = 30240

I also tried doing this using brute force. Our result concur until starting with 3:

6 - 1 (6789T)
5 - 5 (56789, 5678T, 5679T, 5689T, 5789T)
4 - 10 (45678, 45679, 4567T, 45689, 4568T, 4569T, 45789, 4578T, 4579T, 4589T)
3 - 35 (34567, 34568, 34569, 3456T; 34578, 34579, 3457T, 34589, 3458T, 3459T; 34678, 34679, 3467T, 34689, 3468T, 3469T; 34789, 3478T, 3479T; 3489T; 35678, 35679, 3567T, 35689, 3568T; 3569T; 35789, 3578T, 3579T, 3589T, 36789, 3678T, 3679T, 3689T, 3789T)

I'll trust you over me any day.

So we're looking at 35 combinations starting with 3. So it's 1, 5, 10, 35...

Was really hoping working those out would reveal a simple rule for the rest, but...

 

[Debater Slayer]

Isn't this asking how many permutations there are of ten things taken five at a time?

Spoiler: suggested solution

P(n,r)=n! (n−r)! = 10! / 5! = 30240

I also tried doing this using brute force. Our result concur until starting with 3:

6 - 1 (6789T)
5 - 5 (56789, 5678T, 5679T, 5689T, 5789T)
4 - 10 (45678, 45679, 4567T, 45689, 4568T, 4569T, 45789, 4578T, 4579T, 4589T)
3 - 35 (34567, 34568, 34569, 3456T; 34578, 34579, 3457T, 34589, 3458T, 3459T; 34678, 34679, 3467T, 34689, 3468T, 3469T; 34789, 3478T, 3479T; 3489T; 35678, 35679, 3567T, 35689, 3568T; 3569T; 35789, 3578T, 3579T, 3589T, 36789, 3678T, 3679T, 3689T, 3789T)

Spoiler

No, permutation implies that you can change the order, but you can't do that here. There is precisely one way to order 1234 in ascending order, for example, which means a question asking for an ascending sequence of 5 numbers out of 10 is just a combination problem.

This question deceptively looks like a permutation problem because "ascending order" evokes the sense of order typical of permutations, but the fact that ascending order precludes any possibility of multiple arrangements of the same numbers renders it into a combination. The word "ascending" immediately disqualifies permutation.

 

[Debater Slayer]

I'll trust you over me any day.

So we're looking at 35 combinations starting with 3. So it's 1, 5, 10, 35...

Was really hoping working those out would reveal a simple rule for the rest, but...

The answer you responded to is incorrect, so you can keep trying!

 

[Left Coast]

@Polymath257 just tell us the answer.

 

[Debater Slayer]

@Polymath257 just tell us the answer.

No! Don't make him spoil it!

 

[Debater Slayer]

Spoiler: The answer

This is a combination problem without repetition, which means we're calculating the number of ways in which we can arrange r objects out of a total of n objects where order does not matter and none of the chosen objects can be repeated (i.e., we can't use any number more than once in the same arrangement).

The unimportance of order here is the trickiest part: if order mattered, we would use permutations instead. An example of an arrangement where order matters is a competition with three people: an ordered triple where Bob is first, Joe is second, and Max is third is (B, J, M), but changing the order of any of them creates a new ordered triple. In total, there are 3P3 ways in which we can arrange the three contestants, which gives 3 x 2 x 1 = 6 arrangements.

In this question, however, the word "ascending" immediately disqualifies permutation because there is exactly one way to arrange a set of r numbers in ascending order: if I tell you to arrange 1, 2, 3, and 4 in ascending order, you can only produce the sequence 1234. We can't change the order, so this is a combination problem, not a permutation.

We have 10 numbers from 1 to 10, and we need to take 5 each time and arrange them in ascending order. 10C5 is spoken as "10 choose 5," which gives us 252 possible arrangements of 5 numbers in ascending order.