In how many ways can the letters of the word ‘FAILURE’ be arranged so that the consonants may occupy only odd positions? Show 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: 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. How many words can form from failure?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.
How many letters are in the word failure?There are seven letters in the word 'FAILURE' out of which the consonants are- F, L and R. And the vowels are A,E,I and U. There are total 4 odd positions (1st, 3rd, 5th and 7th) and 3 even positions (2nd, 4th and 6th) to fill.
How many words can be formed of the letter in a word failure such that vowels always come together?The number of arrangements of the letters in the word FAILURE, so that vowels are always coming together is. 576.
How many words are in Champions?Do spellers in all grades need to learn all 4,000 words? Nested in Words of the Champions are 450 words, divided into grade levels 1-8, that make up the School Spelling Bee Study List.
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