mmmmmoooooorrrrrrreeeeeeeeeee (1 Viewer)

[1234567]

from the eight letters of the word'FREQUENT', three are taken at random and placed in a line, how many different seuences are possible?

 

[BlackJack]

Is the answer 168? i've forgotten how to do this...

 

[Jellymonsta]

thats how i figured it.
8p3/2!

 

[1234567]

the answer is
didididum!!!!!!!

318!

 

[Jellymonsta]

i dont follow.
im pretty confident i did it the right way. way to knock my confidence out from under me

 

[1234567]

think abt it this way, i dunno how they did it at all

 

[Jellymonsta]

that doesnt help me.
i would like to know how they do it. i dont (and never really have) but i get the feeling that im supposed to...

 

[BlackJack]

It looks like they had all the combinations:
8*7*6
=336
and taken out the instances when there the 2 E's are repeated:
1*1*6 * 3! /2
=18

336-18=318

But what about instances when you could choose either E, which counts as two ways but generate identical results? Who knows...

 

[Jellymonsta]

there is something fiendish about maths
everytime i think ive got something down pat, someone demonstrates that i dont. its not fair i should be able to be consistently competent.
i havent even seen that type of question dealt with like that.
if i ignore it will it go away?

 

[BlackJack]

That... I think the answer's just wrong, until it can prove itself.

 

[spice girl]

The answer IS wrong. Here's the right way:

FREQUENT: there are 7 different letters, and 2 'E's

So there are two types of 3-letter words:
1) Two 'E's and something else:
the "something else" is one of the 6 non-'E's (i.e. F,R,Q,U,N, or T)
and this "something else" can either be in the 1st, 2nd or 3rd position. The other 2 E's fill in the blanks.

So: 6 * 3 = 18

2) Three different letters.
There are 7 different letters to choose from for the first position, then 6 remaining letters for the 2nd, then 5 remaining for the 3rd.

Which E we choose is irrelevant because they are the same.

So: 7*6*5 = 210

Total = 210 + 18 = 228

 

[Dumbarse]

that doesnt seem right either
it could be?
but i would never have done it that way

 

[Jellymonsta]

Somebody write out all the possibilities... Tell us which way is right.
But dont miscount

 

[spice girl]

At my skool, everyone believes me when I give them my answer...you can trust me...

 

[Milly]

I got the same as you spice girl

 

[Dumbarse]

Spicegirl - "At my skool, everyone believes me when I give them my answer...you can trust me..."

haha, alrighty i'll trust u
ego is not a dirty word

 

[1234567]

ok that's from the specimen paper

the model answer goes like this:

total num of diff selection = 8c3 = 56

num of sleections which include both e's = 6
num of selections which dunt include both e's = 50

num of arrangement = 50 x 3! + 6x 3!/2! = 318

which i dunt undertand the answer...

 

[McLake]

Originally posted by 1234567
ok that's from the specimen paper

the model answer goes like this:

total num of diff selection = 8c3 = 56

num of sleections which include both e's = 6
num of selections which dunt include both e's = 50

num of arrangement = 50 x 3! + 6x 3!/2! = 318

which i dunt undertand the answer...

METHOD I:
8C3 is obvious

to include both e's selection is e,e,# and there are 6 hashes. Order is not important (yet).

so no e must be 50 (8C3 - 6)

so arrangments are 50 * 3! (cause can arrange three ways) + 6 x 3!/2! (casue can arrange three ways but 2 of them are same due to e's).

METHOD II:
we have 8 letters to choose from

then 7

then 6

so 8*7*6 = 336

but 6*3 are the same

so no of combo = 336 - 18 = 318

 

[spice girl]

Originally posted by McLake


METHOD I:
8C3 is obvious

to include both e's selection is e,e,# and there are 6 hashes. Order is not important (yet).

so no e must be 50 (8C3 - 6)

so arrangments are 50 * 3! (cause can arrange three ways) + 6 x 3!/2! (casue can arrange three ways but 2 of them are same due to e's).

METHOD II:
we have 8 letters to choose from

then 7

then 6

so 8*7*6 = 336

but 6*3 are the same

so no of combo = 336 - 18 = 318

I'm sorry to say that not only is 8C3 NOT obvious, it's also wrong.

For argument's sake, lets rename the 1st E "E1", and the 2nd E as "E2"

Then there would be 8C3 different ways of choosing 3 letters from the now-distinct 8.

However, now remember that in the real question, both E's are the same. So in all the cases where we only chose one "E" we've double-counted ($)(%)(E1) and ($)(%)(E2) have been considered different sets whereas in fact they are the same (since E1=E2)

NB: $, % are arbitrary letters.

So when there aren't 8 distinct letters (such as FREQUENT), there are less than 8C3 ways of choosing three.

There's a similar flaw in method 2 (we've all double-counted every word with only 1 'E')

To backtrack, we need to subtract every selection that has one 'E', i.e. 1 * 3 * 6 * 5 (one way to choose one 'E', 3 different places to put this 'E', 6 different letters to place the 1st non-'E', 5 different letters to place the 2nd non-'E')

318 - 90 = 228 which is my preferred answer.

Dumbarse: sorry, i dun usually show off - i'm just not used to ppl not believing me...

 

[BlackJack]

Let's go in circles again:

But McLake, what about

8*7*6 right is the general...?
Let's say the first two letters aren't E.
Therefore in the last six, there are two E's.
However, picking either of these generates the same result...
Q, U, E(1)
Q, U, E(2)
two ways, one combination, and it only has one E....
(order'd be important 'cause we're placing it in a line)

Therefore, i support spice girl's answer.

Edit: well, it seems like she got there first.