Counting/ Permuations (1 Viewer)

[Kutay]

Hey i was wondering if anyone knew how to do this question...

From the eight letters of the word FREQUENT, three are taken at random and placed in a line. how many differen't letter sequences are possible???



is the answer like 3! over 8! X 2!

??? thats not sure with i think the answer comes to like 1 / 134400 but that done seems right cause that turns out to like a prob answer cause its b/w 0 and 1... so not to sure i know how to do like find how man sequences like u fot total over the ammount of doubles etc or trpples but not sure with this one thankyou

 

[Jago]

8! / 2! ?

or is it 7! x 2?

probably neither

 

[Kutay]

thankyou very much for clarifying tha up

 

[SaHbEeWaH]

You're not trying to find the probability so you don't need to divide the answer by the total amount of possible combinations

Taking 3 letters at random from 8 letters = ⁸P₃
...divide by 2! because of the recurring E

⁸C₃/2! = 168 possible letter sequences

 

[Kutay]

^{8 [SUP/] 3
hmmm see if this works trying to figure out how to get mathematical symbols on here would make it easier hah}

 

[Dumsum]

Alright. Here's what I'd do.

For 3 letters from the word FREQUENT, situations possible are:
1. three different letters
2. two alike

Firstly, find the number of combinations (even though we are in fact looking for permutations).

1. Three different letters from F, R, E, Q, U, N, T can be done in ⁷C₃ ways. [one E has been left out because in this situation I have defined all three letters to be different]
2. Two Es. This leaves F, R, Q, U, N, T. This can be done in 6 ways.

Now, find the number of different arrangements for each of the combinations. (giving the number of permutations)

1. Number of arrangements = ⁷C₃ x 3! [again, there are always 3 different letters here so it can be done in 3! ways]
2. Number of arrangements = 6 x 3!/2! [here there are always 2 Es so it must be divided by 2!]

So the total number of arrangements of three letters from the word FREQUENT is (⁷C₃ x 3!) + (6 x 3!/2!) = 228


Edit: yes I am aware that I could go straight to the number of permutations for each situation, but doing it this way just makes it more clear in my mind.

 

[grendel]

case 1: 3 different letters = 7 x 6 x 5 =210 (using letterbox technique)
case 2: 2 E's = 6 x 3 (as the there are 6 choices for the third letter and it can be in one of three places)

tf total = 228

 

[Dumsum]

^ much more concise, good stuff. although mine looks a lot bigger than what it actually would be in an answer.

 

[Jago]

oh oops, you were suppose to only take 3...

 

[Kutay]

You're not trying to find the probability so you don't need to divide the answer by the total amount of possible combinations

Taking 3 letters at random from 8 letters = ⁸P₃
...divide by 2! because of the recurring E

⁸C₃/2! = 168 possible letter sequences[/QUOTE

so that incorrect ??

 

[Dumsum]

Yes it is incorrect, because there are certain situations where there aren't 2 Es being chosen so in those cases it wouldn't be necessary to divide by 2!.

 

[tiggerfamilytre]

my old enemy, permutations and combinations...
i think, although i'm unsure, that you have to consider a word with:
- no e's
- 1 e
- 2 e's

I don't think that you can say "3 different letters, therefore 7x6x5 ways" because there are 2 e's in choosing the first letter whereas only 1 of every other letter.

Thus, number of ways is:

for no e's: 6C3x3! = 120
for 1 e: 6C2x2x3! = 180
for 2 e's: 6C1x1x(3!/2!) = 18

where nCr represents n choose r

therefore total number of ways is: 318

I probably made some mistake, though

 

[tiggerfamilytre]

Oh well, thanks.

the battle continues ....