Permutations and Combinations (2 Viewers)

[dunjaaa]

Yeah my bad, I should of just said spaces

 

[seanieg89]

It's 1440.

 

[Squar3root]

Would someone please help this guy through his latest psychotic episode.

you don't see it yet but they're making plans I'm telling you. they're going to take over the world and by the time we realise; it will be too late. pls listen

 

[faisalabdul16]

Another perms and combs question haha, ruse paper (2003)

All the letters of the word ENCUMBRANCE are arranged in a line. Find the total number of arrangements, which contain all the vowels in alphabetical order but separated by at least one consonant.

 

[Squar3root]

Another perms and combs question haha, ruse paper (2003)

All the letters of the word ENCUMBRANCE are arranged in a line. Find the total number of arrangements, which contain all the vowels in alphabetical order but separated by at least one consonant.

I remember doing this question lol. spent like 2hrs on it.

iirc the answer is around 40K or something

 

[obliviousninja]

Another perms and combs question haha, ruse paper (2003)

All the letters of the word ENCUMBRANCE are arranged in a line. Find the total number of arrangements, which contain all the vowels in alphabetical order but separated by at least one consonant.

Looooooool. I remember this question, I surrended at 20

 

[RealiseNothing]

A_E_E_U

Insert three consonants yo

(Let consonants be denoted by C)

_A_C_E_C_E_C_U_

Insert rest of consonants yo

 

[faisalabdul16]

I remember doing this question lol. spent like 2hrs on it.

iirc the answer is around 40K or something

Looooooool. I remember this question, I surrended at 20

This question is so annoying! I cant even find solutions to the paper

 

[braintic]

A_E_E_U

Insert three consonants yo

(Let consonants be denoted by C)

_A_C_E_C_E_C_U_

Insert rest of consonants yo

Ummm .... doesn't that involve double counting?
Because you could insert two consonants next to each other in two ways ... one in the first 'pass', one in the second, then vice versa.

 

[dunjaaa]

I not sure if this is correct, but here's my attempt. So lets instead place the consonants down and leave spaces in between each of them. E.g. _C_C_N_N_M_R_B_. There are 8 potential spots where the we can slot in the 4 vowels. Since the vowels must be in alphabetical order, there is only one way this can happen. Hypothetically say I choose these 4 slots: AC_CENEN_MUR_B_, notice how the condition will always hold true when the spaces are removed. Now the consonants can be arranged in 7!/2!2!. So the total number of arrangements is 8C4 x 7!/2!2! = 88200.

 

[RealiseNothing]

Ummm .... doesn't that involve double counting?
Because you could insert two consonants next to each other in two ways ... one in the first 'pass', one in the second, then vice versa.

You just divide by whatever you have to, I haven't looked at the actual repetition but the method should work.

 

[RealiseNothing]

You could also find the total ways to arrange them in a line, then subtract the amount of times they are next to each other.

This would be fairly easy as far as I can see.

 

[Carrotsticks]

It's 88200.

 

[braintic]

Another perms and combs question haha, ruse paper (2003)

All the letters of the word ENCUMBRANCE are arranged in a line. Find the total number of arrangements, which contain all the vowels in alphabetical order but separated by at least one consonant.

I would love to see an easier method, but ........

Allocate positions to the four vowels (1 to 11).

Total number of selections: 11C4 = 330

Number of selections with 4 vowels together:
Pick the first position of the block (1 to 8)
8 ways

Number of selections with 3 vowels together (but not 4):
(9 ways of picking the first position of the block) times (8 ways of placing the 4th vowel) minus (TWICE 8 ways of getting four vowels in a row ... because you can get 4 in a row in TWO ways - the triple then the single, and vice versa)
56 ways

Number of selections of the vowels occurring in two separated pairs:
(9C2 ways of picking the first position of each pair) minus (8 ways of getting 4 in a row) [you can't get 3 in a row]
28 ways
(We don't double the 8 because picking say positions 2/3 for one pair and 4/5 for the second has NOT been counted twice earlier)

Number of selections with 2 vowels together (but not 3 or 4 or two pair)
(10 ways of picking the first position of the pair) times (9C2 ways of placing the other two vowels) - (THRICE 8 ways of getting 4 in a row ... the pair can come at the start, in the middle or at the end) - (TWICE 56 ways of getting 3 in a row) - (TWICE 28 ways of getting 'two pair')
168 ways
(This time the number of ways of picking two pairs WAS counted twice at the start)

Total no of ways = 330 - 168 - 28 - 56 - 8 = 70


There is only ONE ordering with this choice, so the count remains at 70.


Arrange the consonants in the remaining spaces:
7!/(2!2!) = 1260


Total: 70 times 1260 = 88200


EDIT: I see there IS an easier solution below.

 

[braintic]

I not sure if this is correct, but here's my attempt. So lets instead place the consonants down and leave spaces in between each of them. E.g. _C_C_N_N_M_R_B_. There are 8 potential spots where the we can slot in the 4 vowels. Since the vowels must be in alphabetical order, there is only one way this can happen. Hypothetically say I choose these 4 slots: AC_CENEN_MUR_B_, notice how the condition will always hold true when the spaces are removed. Now the consonants can be arranged in 7!/2!2!. So the total number of arrangements is 8C4 x 7!/2!2! = 88200.

Do you have a similarly easy way of doing Cambridge Year 12 Exercise 10F Q17?

 

[Carrotsticks]

I did mine the exact same way that dunjaaa did it.

 

[braintic]

I did mine the exact same way that dunjaaa did it.

So I should also ask you the question in my previous post.

 

[Carrotsticks]

So I should also ask you the question in my previous post.

Can be done reasonably quickly via inclusion/exclusion, which is essentially the same as cases.

Will think of a faster solution whilst I am out grocery shopping tonight. Thank goodness for Woolworths midnight closing time.

 

[braintic]

Thank goodness for Woolworths midnight closing time.

Pity the bottle shops now close at 10.

 

[dunjaaa]

zzz... can't think of any way besides considering cases and using the inclusion-exclusion principle. I will think about it later though, got an essay to write lol