-
Looking for HSC notes and resources? Check out our Notes & Resources page
combination? (1 Viewer)
- Thread starter 1234567
- Start date
BlackJack
Vertigo!
Is this a 3-letter word, with a least 1 vowel and 1 consonant?
Um, order matters.
So its the difference..
Ways(all) - (ways(all vowels) - ways (all consonants)...
26^3 - 5^3 - 21^3...
8190...
chec P(1 vowel) + P(2 vowels)...
5^2 * 21 + 5 * 21^2 ... multiply by arrangements (*3 in both cases)
8190...
I'm happy...
Um, order matters.
So its the difference..
Ways(all) - (ways(all vowels) - ways (all consonants)...
26^3 - 5^3 - 21^3...
8190...
chec P(1 vowel) + P(2 vowels)...
5^2 * 21 + 5 * 21^2 ... multiply by arrangements (*3 in both cases)
8190...
I'm happy...
BlackJack
Vertigo!
Damn, I could never understand how this works... In a minute...
Ed: wait, what's a word equation?
Ed: wait, what's a word equation?
the word is 'equation'.
ok here is how they did it
there are 5 vowels, so select at least 1 vowel
5c1
there are 3 consonants
so
3c1
then that's at least 1 vowel and 1 consonant, then
we need to choose 1 more letter, from the remaining 6 letters
so it's 6c1
there are 3! ways to arrange these 3 letters
therefore the answer is
5c1 x 3c1 x 6c1 x 3! = 5402
i understand this concept perfectly, except in another trail paper i saw a similar question and i did this in the same way, but the answer is not right...
herei will post the other question.
ok here is how they did it
there are 5 vowels, so select at least 1 vowel
5c1
there are 3 consonants
so
3c1
then that's at least 1 vowel and 1 consonant, then
we need to choose 1 more letter, from the remaining 6 letters
so it's 6c1
there are 3! ways to arrange these 3 letters
therefore the answer is
5c1 x 3c1 x 6c1 x 3! = 5402
i understand this concept perfectly, except in another trail paper i saw a similar question and i did this in the same way, but the answer is not right...
herei will post the other question.
question:
a committe of 3 is choosen from a group of four males and five females. the committe must include at least 1 male and at least 1 female. how many combinations are possible/
in this question, the order doesnt' matter.
from males
4c1, females 5c1, and 7c1 from the rest of the them
then the answer is 4c1 x 5c1 x 7c1 = 140 in the same manner as the question above
but the model answer the paper gave is
4c1 x 5c2 + 4c2x 5c1.
the concept is sort for the same, but the answer is different, they get 70........
that's why it got me confused...
a committe of 3 is choosen from a group of four males and five females. the committe must include at least 1 male and at least 1 female. how many combinations are possible/
in this question, the order doesnt' matter.
from males
4c1, females 5c1, and 7c1 from the rest of the them
then the answer is 4c1 x 5c1 x 7c1 = 140 in the same manner as the question above
but the model answer the paper gave is
4c1 x 5c2 + 4c2x 5c1.
the concept is sort for the same, but the answer is different, they get 70........
that's why it got me confused...
BlackJack
Vertigo!
Oh, I see... it's 'equation'.
The committee is a little different.
if you take 4c1*5c1*7c1, you'll make choosing woman A (in the 5) then woman B (in the rest 7 ) AND woman B then woman A two different ways. Which is not good.
The answer had ways(2 women) + ways(one woman).
The committee is a little different.
if you take 4c1*5c1*7c1, you'll make choosing woman A (in the 5) then woman B (in the rest 7 ) AND woman B then woman A two different ways. Which is not good.
The answer had ways(2 women) + ways(one woman).
Dumbarse
Member
"if you take 4c1*5c1*7c1, you'll make choosing woman A (in the 5) then woman B (in the rest 7 ) AND woman B then woman A two different ways. Which is not good. "
isnt it the same for men but??
why is the answer
ways (2 women) + ways (1 woman)
isnt it the same for men but??
why is the answer
ways (2 women) + ways (1 woman)
Dumbarse
Member
i dont get it, can someone please explain it
how do u get 70>
how do u get 70>
ok here is what u do
in their solution
there are two ways to make up the committe of 3 ppl
2 women + 1 men , and 1 women + 2 men
so tha'ts what they ot
in my way of doing it, somehow the order of teh last two people matters, rememebr when u find the possibility of arranging 2 ppl is 2!?
that's why i got 140.
takes a while to realise,, i am still in the process of awakening...dunno how long that's gonna take though.
in their solution
there are two ways to make up the committe of 3 ppl
2 women + 1 men , and 1 women + 2 men
so tha'ts what they ot
in my way of doing it, somehow the order of teh last two people matters, rememebr when u find the possibility of arranging 2 ppl is 2!?
that's why i got 140.
takes a while to realise,, i am still in the process of awakening...dunno how long that's gonna take though.
BlackJack
Vertigo!
It is the same, that is why the 140 is double the answer.
---
The only possible outcoms are two men and one woman and two women and one man. Three men or three women aren't legal. therefore...
Dumbarse
Member
ohh i get it now
i still hate all sorts of probability, its the worst topic
:axedeath:
i still hate all sorts of probability, its the worst topic
:axedeath:
Or another possible way to do it is:
No of ways with at least 1 girl & at least 1 guy
= [No of total unrestricted ways] - [No of ways for all girls] - No of ways for all guys]
= 9c3 - 5c3 - 4c3
= 70
Anyway, back to the original question about the word "equation", there's a problem regarding 1234567's post, where 1234567 wrote:
"therefore the answer is
5c1 x 3c1 x 6c1 x 3! = 5402"
Actually, 5c1 x 3c1 x 6c1 x 3! = 540 not 5402
Also, 540 is still wrong since the highest amount of possible permutations (seeing as there's no letter repitions in the word "equation") is 8p3 = 336 (without any restricitons at all), and clearly the answer must be less than this. We can see that the unwanted permutations are those where all letters are vowels or all letters are consanants, so we subtract these.
No of total unrestricted permutations of the 3 letter word = 8p3
No of total permutations for the 3 letter word with ALL VOWELS = 5p3
No of total permutations of the 3 letter word with ALL CONSANANTS = 3p3
Therefore, the total number of words that can be made out of these letters containing at least a consanant and at least a vowel is:
8p3 - 5p3 - 3p3
= 270
I'm quite sure that 270 should be the answer. Gee I hate 3U probability, perms, and combos!
No of ways with at least 1 girl & at least 1 guy
= [No of total unrestricted ways] - [No of ways for all girls] - No of ways for all guys]
= 9c3 - 5c3 - 4c3
= 70
Anyway, back to the original question about the word "equation", there's a problem regarding 1234567's post, where 1234567 wrote:
"therefore the answer is
5c1 x 3c1 x 6c1 x 3! = 5402"
Actually, 5c1 x 3c1 x 6c1 x 3! = 540 not 5402
Also, 540 is still wrong since the highest amount of possible permutations (seeing as there's no letter repitions in the word "equation") is 8p3 = 336 (without any restricitons at all), and clearly the answer must be less than this. We can see that the unwanted permutations are those where all letters are vowels or all letters are consanants, so we subtract these.
No of total unrestricted permutations of the 3 letter word = 8p3
No of total permutations for the 3 letter word with ALL VOWELS = 5p3
No of total permutations of the 3 letter word with ALL CONSANANTS = 3p3
Therefore, the total number of words that can be made out of these letters containing at least a consanant and at least a vowel is:
8p3 - 5p3 - 3p3
= 270
I'm quite sure that 270 should be the answer. Gee I hate 3U probability, perms, and combos!
BlackJack
Vertigo!
Grrr... stupid combinations
Oh yeah... I didn't check his stuff...
It's the same reasoning I think.
One vowel from 5
One consonant from 3
and one more.
The old method still counted the combinations twice...
('UQi', U from vowels; i from the rest one arrangement AND 'uQI' with I from vowels and another arrangement...)
but wait, if that's the case , then the answer in the book definitely wrong (again!)
Oh yeah... I didn't check his stuff...
It's the same reasoning I think.
One vowel from 5
One consonant from 3
and one more.
The old method still counted the combinations twice...
('UQi', U from vowels; i from the rest one arrangement AND 'uQI' with I from vowels and another arrangement...)
but wait, if that's the case , then the answer in the book definitely wrong (again!)
Dumbarse
Member
1234567 u keep getting double the answer
there must be a reasoning for this./..
there must be a reasoning for this./..
ok that answer is 540, typo
the answer is from their solution
it's 2002 sydney boys high school trial paper.
have a look of u can .
see that's why ig gets me confused,,,,,,,,
coz the way the got this answer is different from how the solution of the second question worked out!
the answer is from their solution
it's 2002 sydney boys high school trial paper.
have a look of u can .
see that's why ig gets me confused,,,,,,,,
coz the way the got this answer is different from how the solution of the second question worked out!
spice girl
magic mirror
- Joined
- Aug 10, 2002
- Messages
- 785
I got 270 as the answer too. Some sydney boys teacher has some real competence problems.Originally posted by wogboy
I'm quite sure that 270 should be the answer. Gee I hate 3U probability, perms, and combos!
p00_p00
Member
dammit
Fuck math!!!! AGHHHHHHH
Just as i get binomial prob and combos etc, u ppl cant even agree on some questions.......
Fuck math!!!! AGHHHHHHH
Just as i get binomial prob and combos etc, u ppl cant even agree on some questions.......