# Permutations and combinations [Solved!]

**karam** 04 Nov 2016, 20:56

### My question

How many words we can get from the word " gammon " .. please I want to know the style of solution>>> thanks

### Relevant page

### What I've done so far

5×4×3×2+4×3×2

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**karam** 04 Nov 2016, 20:56

How many words we can get from the word " gammon " .. please I want to know the style of solution>>> thanks

5×4×3×2+4×3×2

X

How many words we can get from the word " gammon " .. please I want to know the style of solution>>> thanks

Relevant page <a href="/counting-probability/3-permutations.php">3. Permutations</a> What I've done so far 5×4×3×2+4×3×2

**Murray** 05 Nov 2016, 20:20

Hello Karam

Firstly, it depends what we mean by "word". These are real words:

no

on

am

While these are combinations of letters, but not real words:

mma

omm

namg

Do your "words" need to be real words?

X

Hello Karam Firstly, it depends what we mean by "word". These are real words: no on am While these are combinations of letters, but not real words: mma omm namg Do your "words" need to be real words?

**karam** 06 Nov 2016, 02:36

at first thanks , i need the number of all Different arrangements of the letters from " gammon" .. does not require real meaning >> like from the word " sad" we can get 3×2×1 = 6 words(It is not required to have meaning) ... ....thank you again

X

at first thanks , i need the number of all Different arrangements of the letters from " gammon" .. does not require real meaning >> like from the word " sad" we can get 3×2×1 = 6 words(It is not required to have meaning) ... ....thank you again

**Murray** 06 Nov 2016, 20:39

Using such a definition for "word" means we would actually have 15 "words":

s

a

d

sa

as

sd

ds

ad

da

sad

sda

asd

ads

das

dsa

But actually your more precise request, "the number of all different arrangements of the letters" does imply all letters need to be present in the final arrangement.

Have you looked at Theorem 3 and Example 5 on the "Relevant page" above?

X

Using such a definition for "word" means we would actually have 15 "words": s a d sa as sd ds ad da sad sda asd ads das dsa But actually your more precise request, "the number of all different arrangements of the letters" does imply all letters need to be present in the final arrangement. Have you looked at Theorem 3 and Example 5 on the "Relevant page" above?

**karam** 06 Nov 2016, 21:53

thanks , but i mean all arrangements using all letters

gammon , gaommn, gnmmoa,nogmma, ect.............. i need the number of those arrangements >>> .. thank you .. plz look to this example .. How many different formalities that we can be configured from the letters of "peac".....By using the principle of counting in the permutations we find that: the number of methods for selecting the first letter is 4, the number of the methods for selecting the second letter is three ,the number of the methods for selecting the third letter is two and the number of methods for selecting the fourth letter is 1

Therefore, the number of different formalities is

4 × 3 × 2 ×1 =24

peac , paec , pcea ,....ect>>>we don't care about the meaning ....

X

thanks , but i mean all arrangements using all letters gammon , gaommn, gnmmoa,nogmma, ect.............. i need the number of those arrangements >>> .. thank you .. plz look to this example .. How many different formalities that we can be configured from the letters of "peac".....By using the principle of counting in the permutations we find that: the number of methods for selecting the first letter is 4, the number of the methods for selecting the second letter is three ,the number of the methods for selecting the third letter is two and the number of methods for selecting the fourth letter is 1 Therefore, the number of different formalities is 4 × 3 × 2 ×1 =24 peac , paec , pcea ,....ect>>>we don't care about the meaning ....

**Murray** 07 Nov 2016, 02:53

Hello Karam

1. The first part of my reply was addressing your request for "words" formed from those letters, which was your opening question.

2. The second part of my reply indicated that it's now clearer what you need - arrangements of all the letters in the word "gammon".

3. The last part of my reply asked if you have read the theorem and example (in the link provided above, "3. Permutations") that will help you find your answer.

Please read that first - it will help a lot.

X

Hello Karam 1. The first part of my reply was addressing your request for "words" formed from those letters, which was your opening question. 2. The second part of my reply indicated that it's now clearer what you need - arrangements of all the letters in the word "gammon". 3. The last part of my reply asked if you have read the theorem and example (in the link provided above, "3. Permutations") that will help you find your answer. Please read that first - it will help a lot.

**karam** 07 Nov 2016, 07:29

thank you ,Previously i can't get

"Theorem 3 and Example 5"

then from " gammon" we can get 6!/2! = 360 Various arrangements ...

Do you agree ?

Please accept my apologies

??

??

X

thank you ,Previously i can't get "Theorem 3 and Example 5" then from " gammon" we can get 6!/2! = 360 Various arrangements ... Do you agree ? Please accept my apologies ?? ??

**Murray** 07 Nov 2016, 19:23

Yes, you are correct.

More completely, we have

one g

one a

two m's

one o

one n

So the number of ways to arrange the letters in the word "gammon" is:

`(6!)/(1!xx1!xx2!xx1!xx1!)=720/2=360`

X

Yes, you are correct. More completely, we have one g one a two m's one o one n So the number of ways to arrange the letters in the word "gammon" is: `(6!)/(1!xx1!xx2!xx1!xx1!)=720/2=360`

X

Thanks ..you are the best friend

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