ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: clıppy on November 11, 2012, 11:29:44 am

Title: Quick Combinatorics Question
Post by: clıppy on November 11, 2012, 11:29:44 am: I've got this question that I just can't figure out:

There are 5 vowels and 21 consonants in the English alphabet. How many different four-letter 'words' can be formed that contain two different vowels and two different consonants?

The answer is apparently 11,025 but i got 50,400, can anyone explain this to me?
Title: Re: Quick Combinatorics Question
Post by: b^3 on November 11, 2012, 01:00:17 pm: These 'words' can have the vowels and consonants in any order.
So that is we can have (where v is a vowel and c is a consonant)

$\begin{bmatrix}v\end{bmatrix}\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}v\end{bmatrix}\begin{bmatrix}c\end{bmatrix}$
$\begin{bmatrix}v\end{bmatrix}\begin{bmatrix}v\end{bmatrix}\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}c\end{bmatrix}$
$\begin{bmatrix}v\end{bmatrix}\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}v\end{bmatrix}$
$\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}v\end{bmatrix}\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}v\end{bmatrix}$
$\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}v\end{bmatrix}\begin{bmatrix}v\end{bmatrix}$
$\begin{bmatrix}c\end{bmatrix}\begin{bmatrix}v\end{bmatrix}\begin{bmatrix}v\end{bmatrix}\begin{bmatrix}c\end{bmatrix}$

Now the number of ways we can arrange the first row is
$\begin{bmatrix}5\end{bmatrix}\begin{bmatrix}21\end{bmatrix}\begin{bmatrix}4\end{bmatrix}\begin{bmatrix}20\end{bmatrix}$
We minus 1 the second time because we can't use the same vowel or the sam consonant twice.
If we do the same for the second row we get
$\begin{bmatrix}5\end{bmatrix}\begin{bmatrix}4\end{bmatrix}\begin{bmatrix}21\end{bmatrix}\begin{bmatrix}20\end{bmatrix}$

So as we can see all the rows will have $21\times20\times5\times4$ options, and we have six rows.
So the total number of 'words' that can be formed is $6\times21\times20\times5\times4=50,400$

Thats probably a more intutive way of doing it, but it still gets the answer :)
Title: Re: Quick Combinatorics Question
Post by: clıppy on November 11, 2012, 01:02:53 pm: So the answer is 50,400 not 11,025 making the back of the book wrong. Thanks b^3
Title: Re: Quick Combinatorics Question
Post by: b^3 on November 11, 2012, 01:07:37 pm: Quote from: noclip on November 11, 2012, 01:02:53 pm
So the answer is 50,400 not 11,025 making the back of the book wrong. Thanks b^3
Oh wait... I read what you said wrong before... give me a minute, I'm probably wrong too...... (still waking up...)

EDIT: I think it looks right.... unless someone else can find something wrong with it.

(also we have another westsider on AN! :P)
Title: Re: Quick Combinatorics Question
Post by: clıppy on November 11, 2012, 01:10:02 pm: Haha that's alright
Title: Re: Quick Combinatorics Question
Post by: TrueTears on November 11, 2012, 10:14:47 pm: $4! \binom{5}{2} \binom{21}{2}$

I'll let you think about why