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Question: How many 4 letter words can be formed using the letters of the word FAILURE so that 1) F is includ...

How many 4 letter words can be formed using the letters of the word FAILURE so that

  1. F is included in each word
  2. F is not included in any word
Explanation

Solution

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 6C3^{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 6C3^{6}{{C}_{3}} . Similarly, there are 6C4^{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 6C4^{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 nCr^{n}{{C}_{r}} . Therefore, the number of ways of choosing 3 letters from the remaining 6 letters of the word FAILURE is 6C3^{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
    ⇒4!×6C3\Rightarrow 4!{{\times }^{6}}{{C}_{3}}
    We know that nCr=n!(n−r)!r!^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} . Therefore, the above expression can be expanded as
    ⇒4!×6!(6−3)!3! =4!×6!3!3! \begin{aligned} & \Rightarrow 4!\times \dfrac{6!}{\left( 6-3 \right)!3!} \\\ & =4!\times \dfrac{6!}{3!3!} \\\ \end{aligned}
    We know that n!=n×(n−1)×(n−2)!n!=n\times \left( n-1 \right)\times \left( n-2 \right)! . Therefore, the above expression can be expanded as
    ⇒4×3!×6×5×4×3!3!3!\Rightarrow 4\times 3!\times \dfrac{6\times 5\times 4\times 3!}{3!3!}
    Let us cancel the common terms.
    ⇒4×\requirecancel3!×6×5×4×\requirecancel3!\requirecancel3!\requirecancel3! =4×6×5×4 =480 \begin{aligned} & \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{aligned}
    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 6C4^{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
    ⇒4!×6C4\Rightarrow 4!{{\times }^{6}}{{C}_{4}}
    Let us apply the formula of combination in the above expression.
    ⇒4!×6!(6−4)!4! =4!×6!2!4! \begin{aligned} & \Rightarrow 4!\times \dfrac{6!}{\left( 6-4 \right)!4!} \\\ & =4!\times \dfrac{6!}{2!4!} \\\ \end{aligned}
    We know that n!=n×(n−1)×(n−2)!n!=n\times \left( n-1 \right)\times \left( n-2 \right)! . Therefore, the above expression can be expanded as
    ⇒4!×6×5×4×3×2!2!4!\Rightarrow 4!\times \dfrac{6\times 5\times 4\times 3\times 2!}{2!4!}
    Let us cancel the common terms.
    ⇒\requirecancel4!×6×5×4×3×\requirecancel2!\requirecancel2!\requirecancel4! =6×5×4×3 =360 \begin{aligned} & \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{aligned}
    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 nPr=n!(n−r)!^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} . Students must never miss multiplying 4! With the combination in the above solution.