Permutations and combinations [Solved!]

Re: Permutations and combinations

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?

Re: Permutations and combinations

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

Re: Permutations and combinations

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?

Re: Permutations and combinations

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 ....

Re: Permutations and combinations

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.

Re: Permutations and combinations

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

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Re: Permutations and combinations

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`

Re: Permutations and combinations

karam 07 Nov 2016, 21:03

Thanks ..you are the best friend