# How many different words ending and beginning with a consonant can be made out of the letter of the word EQUATION?

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Hint: First of all, we have a total of 8 places. Fill the first and the last places by 3 consonants in 3 x 2 ways. Now fill the remaining places with the remaining 6 letters in 6! ways. Remember that there is no constraint in filling places from second to seventh place.

Here, we are given a word EQUATION and we have to find the number of words formed by letters of word EQUATION such that the words end and begin with consonants.

First of all, we know that in English language, we have a total of 5 vowels and these vowels are A, E, I, O, U.

Now, the vowels present in the given word EQUATION are E, U, A, I, O. Therefore, the number of vowels present in the word EQUATION = 5.

Also, we know that in English language, all the letters except vowels are known as consonants. We have a total 21 consonants in English.

Now, the consonants present in the given word EQUATION are Q, T, N.

Therefore, the number of consonants present in the word EQUATION = 3.

Therefore, our word has \[\text{ }\]

Now, we have to form 8 letter words such that letters at the beginning and end are consonants.

\[\underline{3}\times \underline{6}\times \underline{5}\times \underline{4}\times \underline{3}\times \underline{2}\times \underline{1}\times \underline{2}\]

We know that we have to fill the first place with consonants only and we have a total of 3 consonants, so we can fill the first place in 3 ways. Also, we have to fill the eighth or the last place with consonants only and we have the remaining 2 consonants, so we can fill the last place in 2 ways.

Now, there are constraints to fill the remaining 6 places with 6 letters. So, any one of the 6 letters can occupy second place. So, there are a total 6 ways to occupy second place.

Likewise, third place can be filled by any of the remaining 5 letters. So, there are a total of 5 ways to occupy third place.

Similarly, fourth place can be filled by any of the remaining 4 letters. So there are a total of 4 ways to occupy fourth place.

In the similar way, fifth, sixth and seventh places can be occupied by remaining 3, 2 and 1 letters respectively. So, there are 3, 2, and 1 ways to fill fifth, sixth and seventh places respectively.

Therefore, we get the total number of words ending and beginning with consonants

\[\begin{align}

& =3\times 6!\times 2 \\

& =3\times 6\times 5\times 4\times 3\times 2\times 1\times 2 \\

& =4320 \\

\end{align}\]

Hence, we can form a total 4320 different words ending and beginning with consonants with the letters of the word EQUATION.

Note: Here, students get confused if repetition is allowed or not. As it is clearly mentioned that the first and last letters of the word are consonants and words should be formed by letters of the word EQUATION then consider that repetition is not allowed. Also, it would not make sense if repetition is allowed as we can make infinitely many words if the repetition is allowed. Also, some students try to do this question manually by writing different words and counting them one by one which is a very lengthy approach and can give wrong results.

Here, we are given a word EQUATION and we have to find the number of words formed by letters of word EQUATION such that the words end and begin with consonants.

First of all, we know that in English language, we have a total of 5 vowels and these vowels are A, E, I, O, U.

Now, the vowels present in the given word EQUATION are E, U, A, I, O. Therefore, the number of vowels present in the word EQUATION = 5.

Also, we know that in English language, all the letters except vowels are known as consonants. We have a total 21 consonants in English.

Now, the consonants present in the given word EQUATION are Q, T, N.

Therefore, the number of consonants present in the word EQUATION = 3.

Therefore, our word has \[\text{ }\]

Now, we have to form 8 letter words such that letters at the beginning and end are consonants.

\[\underline{3}\times \underline{6}\times \underline{5}\times \underline{4}\times \underline{3}\times \underline{2}\times \underline{1}\times \underline{2}\]

We know that we have to fill the first place with consonants only and we have a total of 3 consonants, so we can fill the first place in 3 ways. Also, we have to fill the eighth or the last place with consonants only and we have the remaining 2 consonants, so we can fill the last place in 2 ways.

Now, there are constraints to fill the remaining 6 places with 6 letters. So, any one of the 6 letters can occupy second place. So, there are a total 6 ways to occupy second place.

Likewise, third place can be filled by any of the remaining 5 letters. So, there are a total of 5 ways to occupy third place.

Similarly, fourth place can be filled by any of the remaining 4 letters. So there are a total of 4 ways to occupy fourth place.

In the similar way, fifth, sixth and seventh places can be occupied by remaining 3, 2 and 1 letters respectively. So, there are 3, 2, and 1 ways to fill fifth, sixth and seventh places respectively.

Therefore, we get the total number of words ending and beginning with consonants

\[\begin{align}

& =3\times 6!\times 2 \\

& =3\times 6\times 5\times 4\times 3\times 2\times 1\times 2 \\

& =4320 \\

\end{align}\]

Hence, we can form a total 4320 different words ending and beginning with consonants with the letters of the word EQUATION.

Note: Here, students get confused if repetition is allowed or not. As it is clearly mentioned that the first and last letters of the word are consonants and words should be formed by letters of the word EQUATION then consider that repetition is not allowed. Also, it would not make sense if repetition is allowed as we can make infinitely many words if the repetition is allowed. Also, some students try to do this question manually by writing different words and counting them one by one which is a very lengthy approach and can give wrong results.

Last updated date: 28th Sep 2023

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