How many words can be formed from the word forget if the vowels are together

1). tore 2). fret 3). froe 4). frog 5). reft 6). goer 7). gore 8). grot 9). ogre 10). fort 11). tref 12). rote 13). ergo 14). fore

3 letter Words made out of forget

1). reg 2). toe 3). fog 4). ore 5). tog 6). ort 7). teg 8). ref 9). rot 10). roe 11). ret 12). tor 13). oft 14). eft 15). fet 16). foe 17). for 18). fro 19). get 20). erg 21). gor 22). got 23). ego 24). fer

Real names tell you the story of the things they belong to in my language, in the Old Entish as you might say. It is a lovely language, but it takes a very long time to say anything in it, because we do not say anything in it, unless it is worth taking a long time to say, and to listen to.

I simply regard romantic comedies as a subgenre of sci-fi, in which the world created therein has different rules than my regular human world.

My brother Allie had this left-handed fielder's mitt. He was left-handed. The thing that was descriptive about it, though, was that he had poems written all over the fingers and the pocket and everywhere. In green ink. He wrote them on it so that he'd have something to read when he was in the field and nobody was up at bat.

The word UNIVERSITY consists of 10 letters that include four vowels of which two are same.
Thus, the vowels  can be arranged amongst themselves in

\[\frac{4!}{2!}\]ways.Keeping the vowels as a single entity, we are left with 7 letters, which can be arranged in 7! ways.
By fundamental principle of counting, we get,
Number of words =  7!\[\times\]\[\frac{4!}{2!}\] = 60480

(Having thought through further, this is a shorter solution, and in some ways similar to that posted by true blue anil earlier; the original answer posted is given after this one)

• Assume all letters are different, i.e. no repeats. e.g. the two $N$s are $N_1$ and $N_2$.
• Head: Number of arrangements with consonants only =$^5P_2$.
• Removing arrangements for same base consonants ($N_1 N_2, N_2 N_1$) gives $^5P_2-2$.
• Body: Number of arrangements =$7!$.
• Tail: Number of arrangements with vowels only =$^6P_2$.
• Removing arrangements for same base vowels ($I_1 I_2, I_2I_1, A_1A_2, A_2A_1$) gives $^6P_2-4$.
• Overall: Total number of arrangements, normalising for repeats, is given by

$$\frac {({^5P_2}-2)\cdot 7!\cdot ({^6P_2}-4)}{2!\ 2!\ 2!}=\frac {18\cdot 7!\cdot 26}{8}=294840\;\blacksquare$$

(What follows below is the original answer posted, which is a bit longer and uses a different approach)

We note the following points:

• Because of the way the question is structured, we can focus on the number of ways to construct the Head (2 letters), the Tail (2 letters) and the Body (7 letters).
• There are no repeats in the Head and Tail, so repeats occur - if they do - only in the Body.

Head

• (a) If the Head contains 1 N ($3\times 2=6$ arrangements), then there is $0$ repeat N in the Body
• (b) If the Head contains 0 N ($^3P_2=6$ arrangements), then there is $1$ repeat N in the Body.

Tail

• (1) If the Tail contains one 1 A and 1 I ($2$ arrangements: AI, IA), then there is $0$ repeat A or I in the Body

• (2) If the Tail contains either 1 A or 1 I ($8$ arrangements: $^4P_3-2-2$), then there is $1$ repeat I or A respectively in the Body

• (3) If the Tail contains 0 A and 0 I ($2$ arrangements: EO, OE), then there are $2$ repeats (one each for A and I) in the Body.

Body

• The number of arrangements of letters in the body depends on the number of repeats and is given by $7!/(2!)^n$ where $n$ is the number of repeats in the body for a given Head-Tail combination, contributed by both the Head and the Tail.

Permutations

The number of arrangements or permutations given by different Head-Tail combinations are as follows:

(a) (1): $6\times 2\times {7!}/{(2!)^{0+0}}\color{lightgrey}{=6\times 16\times 7!/(2!)^3}=\;\;60480$
(a) (2): $6\times 8\times {7!}/{(2!)^{0+1}}\color{lightgrey}{=6\times 32\times 7!/(2!)^3}=120960$
(a) (3): $6\times 2\times {7!}/{(2!)^{0+2}}\color{lightgrey}{=6\times \ \ 4 \times 7!/(2!)^3}=\;\;15120$
(b) (1): $6\times 2\times {7!}/{(2!)^{1+0}}\color{lightgrey}{=6\times \ \ 8\times 7!/(2!)^3}=\;\;30240$
(b) (2): $6\times 8\times {7!}/{(2!)^{1+1}}\color{lightgrey}{=6\times 16\times 7!/(2!)^3}=\;\;60480$
(b) (3): $6\times 2\times {7!}/{(2!)^{1+2}}\color{lightgrey}{=6\times \ \ 2 \times 7!/(2!)^3}=\;\;\;\;7560$

Total number of arrangements $\color{lightgrey}{=6\times 78\times 7!/(2!)^3} =294840\;\; \blacksquare$

Complete step by step answer:
We have given 5 vowels and 6 consonants.
Then, we have to find how many 6 letter words can be formed with 3 vowels and 3 consonants.
Now, we need to choose 3 vowels from the given 5 vowels. So, the number of ways to choose vowels will be
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[\begin{align}
  &\Rightarrow {}^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!} \\
 &\Rightarrow {}^{5}{{C}_{3}}=\dfrac{5\times 4\times 3!}{3!\left( 2 \right)!} \\
 &\Rightarrow {}^{5}{{C}_{3}}=\dfrac{5\times 4}{2\times 1} \\
 &\Rightarrow {}^{5}{{C}_{3}}=\dfrac{20}{2} \\
 &\Rightarrow {}^{5}{{C}_{3}}=10 \\
\end{align}\]
Now, we have to find the number of ways to choose 3 consonants from the given 6 consonants, we get
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[\begin{align}
  &\Rightarrow {}^{6}{{C}_{3}}=\dfrac{6!}{3!\left( 6-3 \right)!} \\
 & \Rightarrow {}^{6}{{C}_{3}}=\dfrac{6\times 5\times 4\times 3!}{3!\left( 3 \right)!} \\
 & \Rightarrow {}^{6}{{C}_{3}}=\dfrac{6\times 5\times 4}{3\times 2\times 1} \\
 &\Rightarrow {}^{6}{{C}_{3}}=\dfrac{120}{6} \\
 &\Rightarrow {}^{6}{{C}_{3}}=20 \\
\end{align}\]
Now, we have to find the number of ways to select both vowels and consonants.
We have \[{}^{5}{{C}_{3}}=10\] and \[{}^{6}{{C}_{3}}=20\], so number of ways to choose both will be
\[\begin{align}
  & {}^{5}{{C}_{3}}\times {}^{6}{{C}_{3}}=20\times 10 \\
 & {}^{5}{{C}_{3}}\times {}^{6}{{C}_{3}}=200 \\
\end{align}\]
Now, we have to form 6 letter words by arranging this vowels and consonants. So, 6 letters are arranged in $6!$ ways.
So, the total number of words that can be formed will be
$\begin{align}
  & 6!\times 200 \\
 & =6\times 5\times 4\times 3\times 2\times 1\times 200 \\
 & =144000 \\
\end{align}$

So, total $144000$ words can be formed.

Note: There is a possibility that students may forget to arrange 6 letters to form a word and give the answer as $200$, which is an incorrect answer. Each arrangement of 6 letters gives different words so it is necessary to arrange them within themselves. Also, there is a difference between permutation and combination. Combination means only choosing while permutation means first choosing then arranging.

Which word has all 5 vowels 5 letter word?

How many words can be formed combination vowels are together?

What words have 3 vowels together?

Are there any 5 letter words with 4 vowels?