The number of words of three letters can be formed from the word keppelin

1). peen 2). peep 3). pein 4). peel 5). neep 6). pile 7). pele 8). line 9). pine 10). pipe 11). lien 12). zein 13). plie

3 letter Words made out of zeppelin

1). zip 2). zin 3). pen 4). pep 5). zee 6). pie 7). lei 8). pip 9). pin 10). nil 11). lee 12). lie 13). lin 14). eel 15). lip 16). nee 17). nip 18). pee

List of Words Formed Using Letters of 'zeppelin'

There are 44 words which can be formed using letters of the word 'zeppelin'

2 letter words which can be formed using the letters from 'zeppelin':

el

en

in

li

ne

pe

pi

3 letter words which can be formed using the letters from 'zeppelin':

eel

lee

lei

lez

lie

lin

lip

nee

nil

nip

pee

pen

pep

pie

pin

pip

zee

zin

zip

4 letter words which can be formed using the letters from 'zeppelin':

lien

line

lipe

neep

peel

peen

peep

pein

pele

pile

pine

pipe

plie

zein

6 letter words which can be formed using the letters from 'zeppelin':

lippen

nipple

penile

8 letter words which can be formed using the letters from 'zeppelin':

zeppelin

Other Info & Useful Resources for the Word 'zeppelin'

InfoDetailsPoints in Scrabble for zeppelin21Points in Words with Friends for zeppelin25Number of Letters in zeppelin8More info About zeppelinzeppelinList of Words Starting with zeppelinWords Starting With zeppelinList of Words Ending with zeppelinWords Ending With zeppelinList of Words Containing zeppelinWords Containing zeppelinList of Anagrams of zeppelinAnagrams of zeppelinList of Words Formed by Letters of zeppelinWords Created From zeppelinzeppelin Definition at WiktionaryClick Herezeppelin Definition at Merriam-WebsterClick Herezeppelin Definition at DictionaryClick Herezeppelin Synonyms At ThesaurusClick Herezeppelin Info At WikipediaClick Herezeppelin Search Results on GoogleClick Herezeppelin Search Results on BingClick HereTweets About zeppelin on TwitterClick Here Hint:
Here we will first count the number of letters present in the given word and then we will count the number of vowels and number of consonants present in the word. Then we will take different cases considering the vowels. Then we will find the sum of the number of ways in different cases and solve it further until we will get our required answer.

Complete Step by Step Solution:
Here we need to find the number of different three letters words that can be formed using the letters in the given word i.e. "PROPOSAL".
We can see that there are a total of eight letters in the word "PROPOSAL".
There are three vowels i.e. ‘O’, ‘O’, ‘A’ and there are five consonants i.e. ‘P’, ‘P’, ‘R’, ‘S’, ‘L’.
In first case, we will consider the vowel ‘O’ in the middle i.e. \[\underline {} \underline {\rm{O}} \underline {} \]
The letters left are ‘P’, ‘P’, ‘R’, ‘S’, ‘L’, ‘O’, ‘A’.
As we can see that the letter ‘P’ has occurred two times, we will consider it as one letter. Number of letters left is 6.
Total number of ways to arrange these letters in the first and the third position \[ = {}^6{P_2}\]
Simplifying using the formula\[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], we get
\[ \Rightarrow \] Total number of ways to arrange these letters in the first and the third position \[ = \dfrac{{6!}}{{\left( {6 - 2} \right)!}}\]
Subtracting the terms in the denominator, we get
\[ \Rightarrow \] Total number of ways to arrange these letters in the first and the third position \[ = \dfrac{{6!}}{{4!}}\]
Computing the factorial, we get
\[ \Rightarrow \] Total number of ways to arrange these letters in the first and the third position \[ = \dfrac{{6 \times 5 \times 4!}}{{4!}}\]
Dividing the terms, we get
\[ \Rightarrow \] Total number of ways to arrange these letters in the first and the third position \[ = 6 \times 5 = 30\]
There will be one more word possible which we have not included here i.e. \[\underline P \underline {\rm{O}} \underline P \]
Number of words formed when the vowel ‘O’ is in the middle \[ = 30 + 1 = 31\] ………… \[\left( 1 \right)\]

In second case, we will consider the vowel ‘A’ in the middle i.e. \[\underline {} \underline A \underline {} \]
The letters left are ‘P’, ‘P’, ‘R’, ‘S’, ‘L’, ‘O’, ‘O’.
As we can see that the letter ‘P’ and the letter ‘O’ has occurred two times, we will consider it as one letter. Number of letters left is 5.
Total number of ways to arrange these letters in the first and the third position \[ = {}^5{P_2}\]
Simplifying using the formula\[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], we get
\[ \Rightarrow \] Total number of ways to arrange these letters in the first and the third position \[ = \dfrac{{5!}}{{\left( {5 - 2} \right)!}}\]
Subtracting the terms in the denominator, we get
 \[ \Rightarrow \] Total number of ways to arrange these letters in the first and the third position \[ = \dfrac{{5!}}{{3!}}\]
Computing the factorial, we get
\[ \Rightarrow \] Total number of ways to arrange these letters in the first and the third position \[ = \dfrac{{5 \times 4 \times 3!}}{{3!}}\]
Dividing the terms, we get
\[ \Rightarrow \] Total number of ways to arrange these letters in the first and the third position \[ = 5 \times 4 = 20\]
There will be two more words possible which we have not included here i.e. \[\underline O \underline A \underline O \] and \[\underline P \underline A \underline P \]
Number of words formed when the vowel ‘A’ is in the middle \[ = 20 + 1 + 1 = 22\] ………… \[\left( 2 \right)\]
Hence, the total number of words that can be formed from the given word \[ = 22 + 31 = 53\]

Thus, the correct option is option A.

Note:
Here we need to know the basic formulas of permutation and combination. Permutation is used when we have to find the possible elements but the combination is used when we need to find the number of ways to select a number from the collection. In permutation the order of arrangement matters, whereas in combination, the order of selection, does not matter.
Here, we have to add the possible number of ways to get the required value. If we don't add the possible ways we will not get the correct answer.

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