Question

How many 3 letter words can be formed using the word VOWEL, such that one vowel is always included?

Answer 1

Hint: In this question use the concept if one particular word is fixed then we have to select two more words out of the remaining words so that a three letter word is formed using the combination rule (i.e. if r letters are selected from n letters so the number of ways are (${}^n{C_r}$) so use these concepts to reach the solution of the question.

Complete step-by-step answer:

Given word = VOWEL

As we see there are 5 letters in the given word.

As there are 5 vowel in English alphabet A, E, I, O and U

So only two vowels are present in the word VOWEL which are O and E

Now we have to make 3 letter word such that one vowel is always chosen, so there are following cases arises:

$\left( i \right)$ When O letter is chosen

So we have to choose 2 more letters from the remaining (5 – 1) = 4 letters so that 3 letter word is completed, so we have to choose using the combination rule ${}^4{C_2}$ now we have to arrange these three letters = 3!.

So the number of 3 letter words when O letter is chosen = ${}^4{C_2}$(3!)

 Similarly for rest of the letters

$\left( {ii} \right)$ When E letter is chosen

The number of 3 letter words when E letter is chosen = ${}^4{C_2}$(3!)

Now in both of the above cases both of the two vowels are included in some of the words so we have to consider these words only one time so we have to subtract the two vowel three letter words from these two cases.

So when both of the vowels are selected we have to choose one more letter from the remaining 3 letters = ${}^3{C_1}$, now arrange these letters = 3!

So, the number of 3 letter words when O and E letters are chosen = ${}^3{C_1}$(3!)

So the total number of 3 letter words are the sum of the above two cases and the difference of the third case so we have,

Total number of 3 letter words = ${}^4{C_2}$(3!) + ${}^4{C_2}$(3!) – ${}^3{C_1}$(3!)

Therefore,

= 2(${}^4{C_2}$(3!)) – ${}^3{C_1}$(3!)

Now as we know that \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] so use this property in the above equation we have,

Total number of 3 letter words = \[2.\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}.3! - \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}.3!\]

Now simplify this we have,

Total number of 3 letter words = \[2.\dfrac{{4.3.2!}}{{2.1\left( 2 \right)!}}.3.2.1 - \dfrac{{3.2!}}{{1.2!}}.3.2.1 = 2\left( 6 \right)\left( 6 \right) - 3\left( 6 \right) = 72 - 18 = 54\]

So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that the combination rule of choosing r objects out of n objects and the formula of combination which is stated above, always remember when considering the cases if there are repetition of words in both of the cases so we have to subtract these words one time after the addition of the two cases.