Keeping The Vowels Apart

We need to find the no. of words that can be formed from the letters of the word EXTRA conditioned that the vowels are never together. To do this we follow the follow the following steps:

• STEP 1 : We find the total no. of words that can be formed from the letters of the word EXTRA . No. of such words are:

$=\ ^{5}\mathrm{P}_{5}$

$=5!$

$=120$

• STEP 2 : Then we consider all the words where the vowels appear together ( $E$ and $A$ appear together) and find the no. of such words.( Remember $EA$ can itself be arranged into 2 ways ).Therefore, no. of such words are :

$=\ ^{4}\mathrm{P}_{4} \times \ ^{2}\mathrm{P}_{2}$

$=4! \times 2!$

$=24 \times 2$

$=48$

• STEP 3 : Hence, the desired no. of words or the no. of words that can be formed from the letters of the word EXTRA such that the vowels are never together are : $=120-48$

$=\boxed{72}.$

To find the number of words that vowels "E" and "A" are never together , we can first calculate the total number of possible words formed, and then subtract the number of words that vowels "E" and "A" are together .

$words$ $that$ $vowels$ $not$ $together=total$ $words-words$ $that$ $vowels$ $together$

Let's calculate the total number of words first. Note that the every word has $5$ letters.

There are $5$ possibilities(letters) for the 1st letter, $4$ possibilities for the 2nd letter(because a letter has been used before it, so we need to exclude that letter), $3$ possibillities for the 3rd letter, $2$ possibillities for the 4th letter, and $1$ possibillity for the 5th letter.Total number of words= $5\times4\times3\times2\times1=120$ or $^5P_5=5!=120$ .

Next, we want to find the number words that the vowels are together .

The position of "E" and 'A' must be adjecent, so the possible positions: EA _ _ _ , AE _ _ _ , _ EA _ _ , _ AE _ _ , _ _ EA _ , _ _ AE _ , _ _ _ EA , _ _ _ AE $\rightarrow 8$ possibillities.

How about the remaining $3$ letters(denoted as "_") in the examples? Use the same way as calculating the total of words. $3$ possibillities for the 1st, $2$ possibillities for the 2nd, and $1$ possibilities for the 3rd. $^3P_3=3!=6$ . Number of words that vowels are together= $8\times6=48$

Therefore, the number of words that vowels "E" and "A" are together = $120-48=\boxed{72}$ . Problem solved! :)

5! - (4! x 2) = 72

The number of ways that the vowels can be together is: $5!$ ways because you can choose 1 of the letters without replacement for each position.
If we know how many ways there are the arrange the word so that the vowels "E" and "A" are together, we can find out how many ways they aren't together by subtracting the total number of ways by the ways "E" and "A" are together.
First, clump the letters "E" and "A" together in order to form the string "EA". The next step is to find out how many places the string "EA" can be placed in the word. To do this, the whole string can be treated as one letter. "EA" _ _ _ _ "EA" _ _ _ _ "EA" _ _ _ _"EA"
It is now obvious that there are $4$ positions in total where "EA" can be placed .
Then, find out how many ways there are to arrange the other letters: By the same logic we used to find out how many ways there are to arrange all 5 letters, there are $3!$ ways to arrange the letters "X", "T", and "R" independent of where "EA" is .
Finally, there are $2!$ _ or just _ $2$ ways to arrange the letters inside of the string "EA". ("EA" or "AE")

With these numbers, we can calculate how many ways the letters "EA" are together. The calculation is as follows:
Number of Ways "EA" are Together: $= 4 placements \times \frac{3! "TXR"arrangements}{1 placement} \times \frac{2! "EA" arrangements}{1 placement} = 48 ways$ .

As stated earlier, the total number of ways for "EXTRA" to be arranged is $5! =120$ . The number of ways for "EA" not to be next to each other is the total number of arrangements - the ways "EA" are together.
$120$ _ possible arrangements _ $- 48$ _ ways for "EA" to be together _ $= \boxed{72}$ __ways for "EA" to not be together.

The given word contains 5 different letters. Considering the vowels EA together and treating them as one letter. Thus the letters to be arranged are XTR (EA) - 4 letters. These letters can be arranged in 4! = 24 ways The vowels EA may be arranged amongst themselves in 2! = 2 ways Number of words, each having vowels together = (24x2) = 48 ways Total number of words formed by using all the letters of the given words = 5! = (5x4x3x2x1) = 120 Number of words, each having vowels never together = (120-48) = 72 ................ HOPE ALL LIKE IT........ :-)

First, make the permutations start with $A$ . There are 3 places where $E$ can go, each which has 6 combinations of $R$ , $T$ and $X$ can go ( $3!$ ). By multiplying $3$ by $6$ you get $\boxed{18}$ . Now, we need to find the number of permutations when $A$ is the second letter. There are 2 places where $E$ can go, and once again there are 6 combinations for $R$ , $T$ and $X$ in each so $\boxed{12}$ combinations. When $A$ is the third letter, $E$ can be in 2 places so $\boxed{12}$ combinations. When $A$ is the fourth letter, there are 2 places for $E$ so $\boxed{12}$ combinations. Finally, when $A$ is the 5th letter, $E$ can be in 3 places so $\boxed{18}$ combinations. By adding 18, 12, 12, 12 and 18, you get your answer of $72$ .

2P2 3P3 4=48

5P5=120

120-72=72

You should ask how many"strings" not "words". To most people words have semantic meaning!

Arrangement of Extra = 5! that is nothing but factorial of the number of the character. Let say Ea are together so it can be arrange in two ways. Now Ea together consider one character so now we have 4 place arrangement only means
total number of the arrangement when Ea and together will be = 4! * 2! . so Total Number of arrangement when Ea not together = Total arrangement of the character Without restriction - Total Number of arrangement when Ea are together . and the value is 72

The given word contains 5 different letters. Considering the vowels EA together and treating them as one letter. Thus the letters to be arranged are XTR (EA) - 4 letters. These letters can be arranged in 4! = 24 ways The vowels EA may be arranged amongst themselves in 2! = 2 ways Number of words, each having vowels together = (24x2) = 48 ways Total number of words formed by using all the letters of the given words = 5! = (5x4x3x2x1) = 120 Number of words, each having vowels never together = (120-48) = 72

The given word contains 5 different letters. Considering the vowels EA together and treating them as one letter. Thus the letters to be arranged are XTR (EA) - 4 letters. These letters can be arranged in 4! = 24 ways The vowels EA may be arranged amongst themselves in 2! = 2 ways Number of words, each having vowels together = (24x2) = 48 ways Total number of words formed by using all the letters of the given words = 5! = (5x4x3x2x1) = 120 Number of words, each having vowels never together = (120-48) = 72

Arrange all the consonants as X t r

There are 4 places where vowels can be filled so that they don’t occur simultaneously

No. of ways = 4c2

Total no. of ways = 4c2 * 3! *2! = 72

5!-4!*2=72

The total no of words that can be formed are 5!=120.

Now the total no of words when the letters 'e' and 'a' are together will be 4!=24 and the total no of words when the letters 'a' and 'e' are together will be 4!=24.

Hence the required no of words are 120-(24+24)=72.

ways of arranging the 3 consonants = 3!

ways of putting the 2 vowels in the spaces between the consonants (order matters) = 4P2

for a total of (3!)x(4P2) = 72

5! - 4!*2

In the word EXTRA there are five letters out of which 3 are consonants and 2 are vowels total number of words with letters E,X,T,R,A is 5!=120 consider two vowels (EA) as one letter hence there are four letters X,T,R,(EA) then total number of words are 4!=24 now consider two vowels (AE) as one letter hence there are four letters X,T,R,(AE) then total number of words are 4!=24 hence the total number of words in which A and E are not together are 120-(24+24)=72

The given word contains 5 different letters. Considering the vowels EA together and treating them as one letter. Thus the letters to be arranged are XTR (EA) - 4 letters. These letters can be arranged in 4! = 24 ways The vowels EA may be arranged amongst themselves in 2! = 2 ways Number of words, each having vowels together = (24x2) = 48 ways Total number of words formed by using all the letters of the given words = 5! = (5x4x3x2x1) = 120 Number of words, each having vowels never together = (120-48) = 72

-vipiny35

total words can be formed by extra 5.4.3.2.1.=120 total words as vowels together 4.3.2.1.(2)=48 no of words which vowels are not together 120-48=72

The given word contains 5 different letters. Considering the vowels EA together and treating them as one letter. Thus the letters to be arranged are XTR (EA) - 4 letters. These letters can be arranged in 4! = 24 ways The vowels EA may be arranged amongst themselves in 2! = 2 ways Number of words, each having vowels together = (24x2) = 48 ways Total number of words formed by using all the letters of the given words = 5! = (5x4x3x2x1) = 120 Number of words, each having vowels never together = (120-48) = 72

when vowels are together then the letters can be arranged in 48 ways and total 5 letters can be arranged in 120 ways....therefore, the required answer= 120-48=72

first we will find how many words can be formed by keeping vowel(i.e E & A) together

now assume (E & A) as a bundle..then there will 4 different letters= (E & A), X,T,R..

so total no of words by these 4 letters =4!=24

now (E & A) can be interchanged between them by 2! ways.

so total no. of words in which (E & A) are together= 24*2=48.

since from 5 words total no. of words that can be formed(i.e no. of ways in which 5 words can be arranged)=5!=120.

no. of words in which (E & A) are not together =120-48=72

Firstly, it can be noted that there are $6$ locations at which the vowels can be placed (notated by v):

$v - v - -$ $v - - v -$ $v - - - v$ $- v - v -$ $- v - - v$ $- - v - v$

Because there are two vowels and two positions, there are exactly two permutations to place the vowels in each example, resulting in $12$ total permutations of the vowels.

To place the consonants in any given vowel permutation, we can begin with the $x$ - there are three potential positions from which we can choose. Next, the $t$ is left with two potential positions, and the $r$ must go in the final consonant slot. This leaves: $3! = 3 \times 2 \times 1 = 6$ consonant permutations for each vowel permutation, resulting in $6 \times 12 = \boxed{72}$ total permutations.

1. Anagrams of "EXTRA": $5!$
2. Anagrams such that "E" and "A" be neighbors: $4!2!$
3. By Inclusion-Exclusion: Anagrams of "EXTRA" such that "E" and "A" don't be neighbors: $5! - 4!.2!$

There are 5 letters in the word 'EXTRA' from which we can get 5! different words without any restriction.

Now, if we consider EA as a single block, then we get 4 letters from which we can get 3!*2 words.(multiplying by for for both arrangements AE and EA)

Hence, total words that can be formed from the letters of the word extra, so that the vowels are never together are:

5! - 2*4!=72.

Total number of words=5!=120

now let us consider A and E as one unit

then we have following elements to form a word

(E,A)X,T,R

total number of ways =48

4!*2! (4! for arranging 4 elements and 2! for arranging E and A)

Now 120-48= 72(Required Answer )

Total Number of words which can be formed by rearrangement of letters of 'EXTRA' = 5! = 120. Total Number of words which can be formed by rearrangement of letters of 'EXTRA' where vowels are together(e.g 'EAXTR') can be found out by keeping the two vowels 'E' and 'A' together and treating them as a single letter. Number of such words are = 4! = 24. Also these letter in the vowels group can be arranged in 2! = 2 ways. Hence total number of words that can be formed from the letters of the word EXTRA ,so that the vowels are always together = 24 X 2 = 48. Hence total number of words that can be formed from the letters of the word EXTRA ,so that the vowels are never together is = 120 - 48 = 72.

5!-4!x2=72

Once positions are chosen for the vowels, there are 3! =6 ways to arrange the consonants. There are 5 positions to place E in the 5 letter word. Making it the first or last letter allows 3 possible positions for A. Putting it in the 3 middle positions only allows two possible positions for A. Hence, no. of words =3!(3+3+2+2+2)=72

Total words that can be formed with 5 letters(E,X,T,R,A) will be 5x4x3x2x1=120

Fixing vowels (E,A) together and forming the word with X,T,R , words that can be formed will 4x3x2.also E &A will give 2 combinations. Total words formed with vowels together=4x3x2x2=48

Therefore words formed so that vowels are not together= total words formed- words with vowels together =120-48 =72

5! - 2 * 4! = 72

The total number of words that can be formed using the letters of the word EXTRA is 5!=120. Now, consider the two vowels A and E to be joined together to form a single letter. Now, the total number of words that can be formed using the letters of this new word is 4!=24. However, the two letters A and E can be interchanged in each of these 24 words. Hence, the number of words that can be formed from the letters of the word EXTRA, so that the two vowels ARE TOGETHER is 24*2=48. Hence, the number of words that can be formed so that the two vowels are not together will be= (Total number of words) - (Number of words in which the two vowels are together) = 120-48= 72. Therefore, the required answer is 72.