How many words can be formed from the letters in the word failure the four vowels always coming together?

In how many ways can the letters of the word ‘FAILURE’ be arranged so that the consonants may occupy only odd positions?

Solution:

The word ‘FAILURE’ has four vowels (E, A, I, U)

The number of consonants is three (F, L, R)

Let’s use the letter C to represent consonants.

1, 3, 5, or 7 are the odd spots.

The consonants can be placed in 4P3 ways in these 4 odd spots.

The remaining three even places (2, 4, 6) will be filled by the four vowels. This can be accomplished in a variety of 4P3 methods. As a result, the total number of words with consonants in odd locations = 4P3 × 4P3.

Using the formula, we can

$ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! $

$ P\text{ }\left( 4,\text{ }3 \right)\text{ }\times \text{ }P\text{ }\left( 4,\text{ }3 \right)\text{ }=\text{ }4!/\left( 4-3 \right)!\text{ }\times \text{ }4!/\left( 4-3 \right)! $

$ =\text{ }4\text{ }\times \text{ }3\text{ }\times \text{ }2\text{ }\times \text{ }1\text{ }\times \text{ }4\text{ }\times \text{ }3\text{ }\times \text{ }2\text{ }\times \text{ }1 $

$ =\text{ }24\text{ }\times \text{ }24 $

$ =\text{ }576 $

As a result, there are 576 different ways to arrange the consonants so that they only appear in odd positions.

Answer

Verified

Hint: There are 7 letters in the word FAILURE. Of these, we have to choose 4 letters such that F is present in each word. Therefore, F is fixed. So the number of remaining letters is 3. We can choose these 3 letters from the remaining 6 letters of the word FAILURE in $^{6}{{C}_{3}}$ . These 4 letters can be arranged in 4! ways. To find the number of four letter words that can be formed using the letters of FAILURE so that F is included in each word, we have to multiply 4! and $^{6}{{C}_{3}}$ . Similarly, there are $^{6}{{C}_{4}}$ ways to select 4 letters from the word FAILURE by excluding F. These 4 letters can be arranged in 4! ways. By multiplying 4! and $^{6}{{C}_{4}}$ , we can find the required number of words by excluding F.

Complete step by step answer:

We have to find the number of four letter words that can be formed using the letters of the word FAILURE so that F is included in each word and F is not included in any word. We can see that the total number of letters in the word FAILURE is 7. Of these, we have to choose 4 letters. Let us find the solution to each of the given sections.1) From the 4 letters, we have to include F in each word. Therefore, remaining letters are 3. We can choose 3 letters from the remaining 6 letters of the word FAILURE . We know that the number of ways to selecting r objects from n objects is given by $^{n}{{C}_{r}}$ . Therefore, the number of ways of choosing 3 letters from the remaining 6 letters of the word FAILURE is $^{6}{{C}_{3}}$ . Now, we can arrange the four letters in 4! ways. Therefore, number of four letter words that can be formed using the letters of the word FAILURE so that F is included in each word is given by$\Rightarrow 4!{{\times }^{6}}{{C}_{3}}$We know that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ . Therefore, the above expression can be expanded as$\begin{align} & \Rightarrow 4!\times \dfrac{6!}{\left( 6-3 \right)!3!} \\ & =4!\times \dfrac{6!}{3!3!} \\ \end{align}$We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)!$ . Therefore, the above expression can be expanded as$\Rightarrow 4\times 3!\times \dfrac{6\times 5\times 4\times 3!}{3!3!}$Let us cancel the common terms.$\begin{align} & \Rightarrow 4\times \require{cancel}\cancel{3!}\times \dfrac{6\times 5\times 4\times \require{cancel}\cancel{3!}}{\require{cancel}\cancel{3!}\require{cancel}\cancel{3!}} \\ & =4\times 6\times 5\times 4 \\ & =480 \\ \end{align}$Therefore, there are 480 words that can be formed using the letters of the word FAILURE so that F is included in each word.2) From the 4 letters, we should not include F. Therefore, there are total of 6 letters in the word FAILURE by eliminating F. From these 6 letters, we have to choose 4 letters. We can do this in $^{6}{{C}_{4}}$ ways. We can arrange these 4 letters in 4! Ways. Therefore, the number of four letter words that can be formed using the letters of the word FAILURE so that F is not included in any word is given by$\Rightarrow 4!{{\times }^{6}}{{C}_{4}}$Let us apply the formula of combination in the above expression.$\begin{align} & \Rightarrow 4!\times \dfrac{6!}{\left( 6-4 \right)!4!} \\ & =4!\times \dfrac{6!}{2!4!} \\ \end{align}$We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)!$ . Therefore, the above expression can be expanded as$\Rightarrow 4!\times \dfrac{6\times 5\times 4\times 3\times 2!}{2!4!}$Let us cancel the common terms.$\begin{align} & \Rightarrow \require{cancel}\cancel{4!}\times \dfrac{6\times 5\times 4\times 3\times \require{cancel}\cancel{2!}}{\require{cancel}\cancel{2!}\require{cancel}\cancel{4!}} \\ & =6\times 5\times 4\times 3 \\ & =360 \\ \end{align}$Therefore, there are 360 words that can be formed using the letters of the word FAILURE so that F is not included in any word.

Note: Students must be thorough with the concept and formulas of permutation and combination. They must know when to apply permutation and combination. We use permutation, when the order is a concern while combination doesn’t consider the order of elements. They have a chance of getting confused with the formulas of permutation and combination. The formula for permutation is $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ . Students must never miss multiplying 4! With the combination in the above solution.