hard permutations question (1 Viewer)

[leapordfist]

hi guys, im generally alright in perms and coms, but this one question threw me off
there are 9 numbers
1 2 3 4 5 6 7 8

how many ways can you organise the numbers if the odd numbers are in an increasing order from left to right.

now this is an exam question. if u acaulltly calcualte all the possibilites its very long, as there can be even numbers in between the odd numbers and not even numbers in between.

my guess was that i calcilated the total possibilites and divided them by two. silly guess, yet i just did it haha.

 

[gurmies]

is 9 included too, because you have only 8 numbers there..

 

[Top Secret]

This is why I hate permutations.

 

[ninetypercent]

assuming that the numbers are 1 2 3 4 5 6 7 8 9
ignoring the even numbers. There are five odd numbers

11111 _ _ _ _

1 can go in 5 places
same with 3, 5, 7 and 9.

5^5 = 3125

 

[gurmies]

assuming that the numbers are 1 2 3 4 5 6 7 8 9
ignoring the even numbers. There are five odd numbers

11111 _ _ _ _

1 can go in 5 places
same with 3, 5, 7 and 9.

5^5 = 3125

Not quite that simple I think.

 

[LordPc]

since the odd numbers are in increasing order, and so can only be ordered in one way, they can be generalised to sticks

so you must now sort 4 sticks and 4 numbers, 2, 4, 6, and 8
eg |2|4|6|8 = 1 2 3 4 5 6 7 8

so how many ways can you arrange the sticks? I believe it is 8C4 = 70 (8 items altogether, choose 4, order unimportant)

then you just arrange the remaning 4 numbers, which are 2, 4, 6, 8 which is 4!

so the answer is 4! * 70 = 4*3*2*70 = 280*3*2 = 560*3 = 1500 + 180 = 1680

correct?

(side note: 8P4 = 1680, so perhaps there is a different appoach that would be faster)

 

[untouchablecuz]

is it 8P4

 

[lychnobity]

hi guys, im generally alright in perms and coms, but this one question threw me off
there are 9 numbers
1 2 3 4 5 6 7 8

how many ways can you organise the numbers if the odd numbers are in an increasing order from left to right.

now this is an exam question. if u acaulltly calcualte all the possibilites its very long, as there can be even numbers in between the odd numbers and not even numbers in between.

my guess was that i calcilated the total possibilites and divided them by two. silly guess, yet i just did it haha.

Is the answer (if the question includes 9):

6C4 x 4! (actually same thing as 6P4 now that I think about it)

ie, only 1 way to arrange the odds, so you must sort the evens. _1_3_5_7_9_

As shown by the _, there are 6 spaces to place the evens, and to order them, it would be 6P4. Or, there are 6 spaces in which you choose 4 (4 numbers to sort remember), and 4! to arrange the even numbers.

 

[gurmies]

Is the answer (if the question includes 9):

6C4 x 4! (actually same thing as 6P4 now that I think about it)

ie, only 1 way to arrange the odds, so you must sort the evens. _1_3_5_7_9_

As shown by the _, there are 6 spaces to place the evens, and to order them, it would be 6P4. Or, there are 6 spaces in which you choose 4 (4 numbers to sort remember), and 4! to arrange the even numbers.

Your method fails to acknowledge that 135792468 is a possibility...as is something like, say, 821345796

 

[Timothy.Siu]

i haven't read other people solution yet,

but is it 4!x5x6x7x8x9 / 5!

 

[Michaelmoo]

You have 5 odd numbers. Now since they can only be arranged in the order 1,3,5,7,9; then you select 5 places for the 9 [i.e. 9C5]. Now theres 4 remaining spots for the even numbers. SInce they dont have to be in any order, they can be arranged in 4! ways. So the solution:

= 9C5 x 4!

= 3024

 

[yibbon]

You have 5 odd numbers. Now since they can only be arranged in the order 1,3,5,7,9; then you select 5 places for the 9 [i.e. 9C5]. Now theres 4 remaining spots for the even numbers. SInce they dont have to be in any order, they can be arranged in 4! ways. So the solution:

= 9C5 x 4!

= 3024

This wont work sorry, just like gurmies said
"Your method fails to acknowledge that 135792468 is a possibility...as is something like, say, 821345796"

I might be wrong but:
[4^5 + 3^5 + 2^5 + 1^5]*4! = 31,200

Chose a place for 1, there are 4 spots as you have to allow room for all the potential spacings. Then there are 4 spots for 3 and so on (hence 4^5)...but I think my logic breaks down as for each 1-4 extra spaces you allow it creates even MORE 1-4 cases.

There are similar alphabetical order questions in Cambridge 3U using words, none of which I have solved either.

 

[ninetypercent]

Yibbon is right. There are much more cases that have to be considered.
This question is so confusing.

hopefully someone here will provide something like a textbook explanation to show us how to do it. I'm scared that this will appear in my exam. It better NOT!

 

[ninetypercent]

i haven't read other people solution yet,

but is it 4!x5x6x7x8x9 / 5!

care to explain Timothy?

 

[yibbon]

A friend brute forced it using a computer program and got 3024. Now im very confused!

 

[lolokay]

it's just 9!/5!

9! to arrange the numbers, then divide by 5! -> having the odd numbers in a given order is basically the same as them all being the same number (i.e. same number of arrangements as total arrangements for 1 1 1 1 1 2 4 6 8)


michaelmoo's reasoning is correct too (pick the spot for the odd, arrange the evens in the other spots -> 9C5*4!)

 

[Timothy.Siu]

it's just 9!/5!

9! to arrange the numbers, then divide by 5! -> having the odd numbers in a given order is basically the same as them all being the same number (i.e. same number of arrangements as total arrangements for 1 1 1 1 1 2 4 6 8)

yeah, i knew it was simpler than what i did (although it only took like 10 seconds)

 

[Drongoski]

I thought it was . . . !

 

[LordPc]

since the odd numbers are in increasing order, and so can only be ordered in one way, they can be generalised to sticks

so you must now sort 4 sticks and 4 numbers, 2, 4, 6, and 8
eg |2|4|6|8 = 1 2 3 4 5 6 7 8

so how many ways can you arrange the sticks? I believe it is 8C4 = 70 (8 items altogether, choose 4, order unimportant)

then you just arrange the remaning 4 numbers, which are 2, 4, 6, 8 which is 4!

so the answer is 4! * 70 = 4*3*2*70 = 280*3*2 = 560*3 = 1500 + 180 = 1680

correct?

(side note: 8P4 = 1680, so perhaps there is a different appoach that would be faster)

ok, since I did it with the numbers 1-8, and you perhaps just went straight to my answer of 1680 and thought i did it wrong, I will explain again using the numbers 1-9

---

since the odd numbers are in increasing order, and so can only be ordered in one way, they can be generalised to sticks

so you must now sort 5 sticks and 4 numbers, 2, 4, 6, and 8
eg |2|4|6|8| = 1 2 3 4 5 6 7 8 9

so how many ways can you arrange the sticks? it is 9C4 = 126 (9 items altogether, choose 4 spots, order unimportant)

then you just arrange the remaning 4 numbers, which are 2, 4, 6, 8 which is 4!

so the answer is 4! * 126 = 3024

---

and could I have a look at that program? I'd like to see how you would brute force this

 

[lolokay]

yeah, i knew it was simpler than what i did (although it only took like 10 seconds)

if you started with the odd numbers, you could just do 6*7*8*9 to put the evens in