Question

A challenging permutation question
@Miracrown @ganeshie8 @inkyvoyd @mayankdevnani @phi @mathmale
@Directrix @hartnn @zepdrix @aaronq @kropot72 @ParthKohli

Answer 1

How many different 4 or 5 letter words can be formed from TELEPHONEL

Answer 2

@FaiqRaees is the answer 35280 ??? :/

Answer 3

Would have been easier if it were combinations. lol

Answer 4

I wonder if this is really a challenge or just someone's homework question they need help with.

Answer 5

Can the letter in the 4/5 letter words be equal? (e.g. E.E.P.H.)

Answer 6

@rishavraj Answer's 8130
@Hero Trust me not a homework question.
@reemi I couldn't understand your question

Answer 7

ok few things. ....its TELEPHONEL ..... ???? and also can the alphabets repeat ??

Answer 8

@FaiqRaees is EEELL a valid word?

Answer 9

@rishavraj Yes and No
@reemi YEs

Answer 10

from @rishavraj : `alphabets can repeat` -> `No`
from @reemii : `is EEELL a valid word?` -> `YEs`.
There's something I don't understand. Can you explain what can repeat and what cannot repeat?

Answer 11

@reemii You can just use the letters given.

Answer 12

EEELL : letters repeat, you said it is ok.
What do you mean by "alphabets cannot repeat"?

Answer 13

Like you cant use T more than once

Answer 14

got it

Answer 15

TTONE is not a valid word

Answer 16

You meant TELEPHONEL, right? Not TELEPHONE I hope.

Answer 17

Yep

Answer 18

T, EEE, LL, O, H, N, P

FOUR LETTER WORDS
Case 1: Three same, one different.
In that case, the three same will be E. The last letter can be chosen in \(6\) ways and its place in the word can be chosen in \(4\) ways. Total words here = 24.
Case 2: Two same, two same.
Here, we've got EELL being permuted so \(4!/2!2! = 6\) words here.
Case 3: Two same, two different.
For two same, we can either go for E or L. \(2\) ways.
Then for the two different, we've got to go for two out of six remaining choices of letters. \(\binom{6} 2\). Select the places for those two letters in \(\binom{4}2\) ways and multiply by \(2!\) to permute. So here, we've got \(2\cdot 2 \cdot \binom{4}2 \binom{6}2=360\).
Case 4: All different.
This one will simply be \(\binom{7}4 4!=840\).

Total no. of four letter words = \(24+6+360+840=1230\).

Answer 19

And about five letter words?

Answer 20

Yeah, I didn't want to fit all that in one post.

Answer 21

There's a little formula, I think, but I obtain 6610 with it.
\[
\sum_{L=4}^5 \sum_{n_E=0}^3 \sum_{n_L=0}^2 \binom{5}{L-n_E-n_L} \frac{L!}{n_E! n_L!}
\]

For \(L=4\) (length of the words),
\[
\sum_{n_E=0}^3 \sum_{n_L=0}^2 \binom{5}{L-n_E-n_L} \frac{L!}{n_E! n_L!} = 1230.
\]

Answer 22

For \(L=5\), I obtain 5380.

Answer 23

Total should be 8130

Answer 24

I don't know. Even this little snippet (Python) convinces me that it's 6610..
```
>>> for p in permutations("TELEPHONEL"):
s.add(p[:5])
s.add(p[:4])

>>> len(s)
6610
```

Answer 25

`s = set()` (before the loop)

Answer 26

OMG I lost whatever I'd typed -_-

Answer 27

AAAAAAh

Answer 28

Just give me the working for each stage I will understand myself

Answer 29

Now I'm gonna do it one at a time.

FIVE LETTER WORDS

T, EEE, LL, O, H, N, P

Case 1: Three same, two same.\[\frac{5!}{3!2!}=10\]

Answer 30

ok

Answer 31

Case 2: Three same, two different.\[\frac{5!}{3!}\binom{6}{2} = 300\]

Answer 32

The term \(\binom{5}{L-n_E-n_L}\frac{L!}{n_E! n_L!}\) means:
1) if there are \(n_E\) letters \(E\) and \(n_L\) letters \(L\) in the word, then one needs to complete the \(L\)-letter word with \(L-n_E-n_L\) letters -> choose as many from a set of 5 different letters.
2) now you have \(L\) letters at hand. There are \(L!\) permutations of those letters, and there might be same elements because there might be several \(E\)'s and several \(L\)'s. Divide by \(n_E!\) and by \(n_L!\).

Answer 33

Case 3: Two same, two same, one different.\[\frac{5!}{2!2!}\cdot 6=180\]

Answer 34

Two same3 different = 1200?

Answer 35

None same = 2520?

Answer 36

Was that all @ParthKohli

Answer 37

Case 4: Two same, three different.\[\frac{5!}{2!}\binom{2}2 \binom{6}{3}=2400\]Case 5: All different.\[\binom{7}55!=2520\]

Answer 38

How you're getting 8130 is something I've got no idea about.

Answer 39

The answer's 8130 in the book

Answer 40

Did we at least get the four-letter part right?

Answer 41

https://www.quora.com/In-how-many-ways-can-5-letters-be-taken-from-the-word-Mississippi
https://www.quora.com/How-many-4-letter-words-can-be-formed-by-the-word-MISSISSIPPI

Answer 42

Dont know @ParthKohli

Answer 43

@ParthKohli Your Case 3 should be \(\frac{5!}{2!2!}5\) because there are \(5\) letters remaining that are neither E or L.

Answer 44

When you sum your separate cases with this correction, you obtain my answer (and the computer's answer) : 6610.

Answer 45

@FaiqRaees

Answer 46

@reemi @ParthKohli Your Case 3 should be \(\frac{5!}{2!2!}5\) because there are \(5\) letters remaining that are neither E or L.
If they are neither E nor L then it should be 5! = 120

Answer 47

@reemi they have already been included in the distinct cases

Answer 48

This "Case 3" deals with the words "2 same, 2 same, 1 different".
The "2 same" and "2 same" must be E,E and L,L. Therefore, the "1 different" must be a letter from the 5 others. -> choose 1 letter from a set of 5 letters. -> 5 choices.
-> permute: 5!
-> divide by 2! * 2!.
-> number of such words \(\frac{5!}{2!2!}\times 5 = 150\).

Answer 49

Oh that one

Answer 50

\[
\underbrace{\frac{5!}{2!2!}}_{\text{different permutations of the word}}\times \underbrace{5}_{\text{number of ways to make a choice of letters to verify "Case 3"}}
\]

Answer 51

so, if you correct 180 to 150, the big total is 6610.

Answer 52

Case 2: Three same, two different.
"Three same" must be "three \(E\)".
Two different: the letters can be from {T,P,H,O,N} and {L} : it's okay if L appears once or not at all. -> {T,P,H,O,N,L} has 6 six letters.

-> 3 \(E\) 's, 2 other letters : there are \(\binom{6}{2}\) ways to choose them.
-> different permutations of the set of letters chosen: \(5! / 3!\) because ther eare 3 \(E\) 's.