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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 listenWould someone please help this guy through his latest psychotic episode.
I remember doing this question lol. spent like 2hrs on it.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 20Another 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
This question is so annoying! I cant even find solutions to the paperLooooooool. I remember this question, I surrended at 20
Ummm .... doesn't that involve double counting?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
You just divide by whatever you have to, I haven't looked at the actual repetition but the method should work.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.
I would love to see an easier method, but ........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.
Do you have a similarly easy way of doing Cambridge Year 12 Exercise 10F Q17?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.
So I should also ask you the question in my previous post.I did mine the exact same way that dunjaaa did it.
Can be done reasonably quickly via inclusion/exclusion, which is essentially the same as cases.So I should also ask you the question in my previous post.
Pity the bottle shops now close at 10.Thank goodness for Woolworths midnight closing time.