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combination? (1 Viewer)

underthesun

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how about a thorough explanation on the subject :D

for the word EQUATION

EUAIO QTN we get styles:

ways(2vowels) * ways(1consonant) because order is important..
+
ways(1consonant) * ways(2vowels) we repeat this again different way
+
ways(2consonant) * ways(1vowels) same thing..
+
ways(1vowels) * ways(2consonant)

maybe that explains why the answer doubles from 270?
 
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you idiots!!!

5402? 540? WHAT THE HELL!!!! (original problem)

firstly - there are 8 letters in 'equation' and youre making 3 letter words, therefore the total possibilities (not excluding anything) is

total ways = 8P3 = 336

therefore anything over 336 is obviously wrong!!!

it was a good idea to get the combinations first and then multiply by 3! for order so lets stick with that...

(method 1) if you do the following...

5c1 x 3c1 x 6c1 then its wrong!!!
(vowel) (consantant) (other)

why? because you are counting combinations twice for example, equ and uqe are counted as separate combinations (we havent considered order yet - we multiply by 3! later)

(method 2) because you cant do the above, we have to take each case separately...

exactly 1 vowel -
5c1 x 3c2 = 15
(vowel) (consanants)

exactly 2 vowels -
5c2 x 3c1 = 30
(vowel) (consanants)

therefore there are (15) + (30) 3 letter combinations which makes (45) 3 letter combinations and multiply this by 3! for the number of possible words

45 x 3! = 270

the answer is 270

you can check it because

(0 vowels) + (1 or 2 v) + (3 v) = total
3P3 + 270 + 5P3 = 8P3
6 + 270 + 60 = 336

thus 270 is correct!!!
 
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you idiots pt II !!!

as for the 2nd problem,

same strategy...

ways 2 men...
4C2 x 5C1 = 30

ways 1 man...
4C1 x 5C2 = 40

therefore: ways(at least 1 guy, 1 chick) = 30 + 40 = 70

thats the answer, 70! you dont double it because order doesnt count!!!

(2C2) the way of arranging 2 people without order is 1
(2P2) the way of arranging 2 people with order is 2
(nCn) the way of arranging n people without order is 1
(nPn) the way of arranging n people with order is n!
 

Dumbarse

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Originally posted by 1234567
the word equation contains five vowels, how many 3 letter words cnsisting of at least 1 vowel and i consonant can be made from the letters of equation?
ok, can someone tell me what i'm doing wrong please!??

equation 5 vowels 3 consonants

1vowel * 1consonat * any other letter

5*3*6 = 90 way

now these 3 letters can be arranged 3*2*1 ways = 6

so 90*6 = 540 ways
 

McLake

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Re: Re: combination?

Originally posted by Dumbarse


ok, can someone tell me what i'm doing wrong please!??

equation 5 vowels 3 consonants

1vowel * 1consonat * any other letter

5*3*6 = 90 way

now these 3 letters can be arranged 3*2*1 ways = 6

so 90*6 = 540 ways
Your "any other letter" has already been counted as a constant or vowel, meaning you've counted the combo twice.

So, divide by 2!
 

Dumbarse

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Re: Re: Re: combination?

Originally posted by McLake


Your "any other letter" has already been counted as a constant or vowel, meaning you've counted the combo twice.

So, divide by 2!
how has it?? please explain

so what does that mean in terms of the 3 letter word>
and why divide by 2!
 

McLake

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Re: Re: Re: Re: combination? <-- getting very long

Originally posted by Dumbarse


how has it?? please explain

so what does that mean in terms of the 3 letter word>
and why divide by 2!
OK, say I picked EXC, then I picked ECX
They're the same! (don't look at order just yet, that a different issue)

Your method counts both of these, right. So every time you pick VCL (vowel, consenet, letter) you've also picked either VCV or VCC. In other words, you've counted things twice.

Do you follow me?
 
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