Thread: Counting Methods

Please help!


1.) Three-letter 'words' are to be made by arranging the letters of the word METHODS. What is the probability that the word begins with a vowel?

I tried to do this question by finding the total number of possibilities then using the formula Pr(event) = Relevant/Total. I did not know how to successfully represent 1 vowel and i did it like this: 7p1 or n English, 1 letter picked out of 7, is this correct?

2.) 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?

For this question i attempted to use permutative formula and got the following: 5p2 x 21p2 and found that it was 8400 ways, it was ultimately incorrect, could someone please explain?

Thanks!

Originally Posted by andrew2322

Please help!


1.) Three-letter 'words' are to be made by arranging the letters of the word METHODS. What is the probability that the word begins with a vowel?

Lets construct a diagram

_ _ _ are the spaces we need to fill to make a "word".

For the first blank we can choose any of the seven letters. The second blank, any of the six remaining letters. The last one, any of the five remaining letters. Therefore, the total number of possibilities is 7 * 6 * 5.

Now let us see how many combinations there are when the first letter must be a vowel. Out of the seven letters in METHODS, only 2 are vowels, so the first blank, choose any of the two vowels. For the second blank, choose any of the six remaining letters. For the third blank, any of your five remaining letters. Total number of combinations then is 2 * 6 * 5

(2*6*5)/(7*6*5) = 2/7

working on the next problem.

Originally Posted by andrew2322

Please help!


1.) Three-letter 'words' are to be made by arranging the letters of the word METHODS. What is the probability that the word begins with a vowel?

The 'word' can start with either 'E' or 'O' .

Number of possible words formed = 2(6x5) =60

Total number of possible words formed =7x6x5 =210

Thus , probability = 60/210

Hello, thanks for the help but i'm still a little confused with this whole counting methods thing.

How do i represent 2 out of 7 letters mathematically, is it like 7p2?

And how do i know when to use permutations instead of combinations? Its sooooo confusing!

Originally Posted by andrew2322

2.) 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?

For this, I constructed it this way. How many unique combinations can I have with two vowels and two consonants in a word? Well, where V is vowel and C is consonant

V V L L
V L V L
V L L V
L L V V
L V L V
L V V L
(verify)

each of these rows each have 5 * 4 * 21 * 20 unique combinations, so
6 (5 * 4 * 21 * 20) = ANS.

EDIT: I would really need to refresh my memory on permutations and combinations, but permutations do not include order and combinations do

hey i got something similar to that except mine was 4!(4*5*21*20

either way, the book says the answer is 11 025.

Yes. I understand that permutations involve order and combinations do not but i don't get it, like some questions that don't involve order use permutations!

Originally Posted by andrew2322

hey i got something similar to that except mine was 4!(4*5*21*20

either way, the book says the answer is 11 025.

hmm, I either missed something crucial in the problem or my construction was incorrect or your book is wrong.

This method of communication is a little inconvenient, is there an easier way to communicate with you?

Originally Posted by andrew2322

[snip]
2.) 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?

[snip]

Won't it be ....?

(Plato will probably correct me here if I'm wrong here ....)

Originally Posted by andrew2322

This method of communication is a little inconvenient, is there an easier way to communicate with you?

It is better to post in the forums than start sending pms for help (MHF is not a tutorial service ....)

Originally Posted by andrew2322

And how do i know when to use permutations instead of combinations? Its sooooo confusing!

Correct me if I am wrong here, but i believe that we use a permutation when the order counts and a combination when it does not.

or vica versa, think i was right the first time though :P

that's right. order is important in permutation.