Question

How many 4 letter arrangements can be made from the letters of the word ELEMENTARY?

Answer 1

true?

Answer 2

Enter
Ment
Elem
Lent

Answer 3

this isn't a text twist game...

Answer 4

this is a permutation question

Answer 5

Oh... My bad.

Answer 6

how many unique letters in "ELEMENTARY"

how many ways can you choose 4 out of the total number of unique letters?

Order does matter, since it's "arrangements"

Answer 7

so you're saying i do 7P4? it doesn't work...

Answer 8

wait. i think i made a calculation error

Answer 9

it's not 11P4/3! either

Answer 10

ELEMENTARY has 10 letters...
1234567890

and 7 unique letters: E, L, M, N, T, A, R, Y

So it's "7 choose 4" = 7! / (7-4)! = 7! / 3! = 840

Answer 11

Arrggh!! i can't count to 7 or 8

Answer 12

hmm my bad...but it really is not 10

Answer 13

i mean 840

Answer 14

9P4 = 9! / (9-4)! = 3024

Answer 15

still no...but i think i have an idea

Answer 16

holy #!*% what is wrong with me...

Answer 17

Maybe 3rd time is the charm...

Answer 18

there are 8 letters. order matters. pick 4 from 8.

8 x 7 x 6 x 5

or 8! / (8-4)!

= 1680

Answer 19

That was very embarrassing... I may need more sleep.

Answer 20

no..still not 1680

Answer 21

Hm. Ok, maybe I don't actually know how to do this... Are you quizzing the community as a teachable moment (I hope) or asking for help (in which case I'm apparently not it)?

Answer 22

no..im just trying as hard as you are

Answer 23

Ok, just checking... I'm "big" enough to admit I don't know... was ready to learn! Guess we'll have to figure it out together :)

Answer 24

Ohh... inspiration...

Answer 25

It isn't unique letters... it's ALL letters.

But the non-unique letters produce duplicate arrangements

Answer 26

So it's 10 choose 4, eliminate duplicates.

So EEEM is valid, just as an example... my earlier math totally missed this type of valid case

Answer 27

7C3 * 4! + 7C2 * 2! + 7C1 * 4 <--too small

Answer 28

I'm rusty at this... how did you get that expression, even if it isn't correct?

Answer 29

7C3 * 4! is because you have 1 E and then you pick 3 out of the other 7 non-E letters.. then you can rearrange it in 4! ways

7C2 is because you pick 2 E and you pick 2 non-E...then rearrange it in 4!/2!

7C1 is because you pick 3 E and 1 non-E..then rearrange it in 4!/3! ways

Answer 30

when you say "arrange it" are you saying arrange the whole 4 letter set? Or just the extra 3, 2, or 1 letters beyond the 1, 2, and 3 E's, respectively?

Answer 31

if i add 7C4 * 4! it becomes closer to the answer

Answer 32

Those E choices seem intended to eliminate duplicates... trying to think through whether you inadvertently "forced" the E choice to be the leading character or whether what you wrote allows, for example, for the 3 E choice to include _EEE

Answer 33

even if i add 7C0 it still doesn't reach the answer

Answer 34

wow.. this is deceptively hard.

Answer 35

indeed

Answer 36

so u know that elementry has 10 letters

Answer 37

yeah

Answer 38

nm. someone already did that

Answer 39

There are 10 P 10 or 10! ways to arrange the letters...but 3 of them are repeats

So you have 3! = 6 sets of duplicates which you have to account for

Answer 40

10!/3! ?

Answer 41

that's rearranging the whole thing...

Answer 42

oh wait, you want 4 letters not 10

so it's 10 P 4 over 3!

Answer 43

yes i tried that...it doesn't work sadly

Answer 44

what did you get

Answer 45

oh i see above, nvm

Answer 46

well if order doesn't matter, then you have to divide that 840 by 4! or 24 to get 35

Answer 47

but there's probably a typo somewhere

Answer 48

that got even smaller

Answer 49

well it's either 840 or 35...or there's a typo somewhere (which is what I'm betting is the case)

Answer 50

no typo and it's neither 840 nor 35

Answer 51

you're not considering the cases of E's

Answer 52

I did

Answer 53

those are the 3 repeats I referred to

Answer 54

you're just considering 1 E and 3 unique letters

Answer 55

no, when I divided by 3!, I'm dealing with the repeated E's

Answer 56

like I said, there's a typo somewhere (not your fault, a typo made by the system or the professor)

Answer 57

wait... you considered 3 E's and 1 unique letter..if you look above i considered 1 E and 3 unique + 2 E's and 2 unique + 3 E's and 1 unique then 4 E's

Answer 58

but it still smaller

Answer 59

anyway the answer should be 1960..that's too far to be a typo

Answer 60

hmm idk how they're getting 1960, the answer should be 840

Answer 61

it's cases of E's

Answer 62

@tcarroll010 perhaps you'll like this