Both of your answers are incorrect. Method 1: We arrange the consonants, then place the vowels in the spaces between and at the ends of the row. There are $4!$ ways of arranging the four distinct consonants B, G, L, R. For each such arrangement, there are five spaces in which we could place the vowels, three between successive consonants and two at the ends of the row. To separate the vowels, we must place the three vowels in three of these five spaces. Choose two of the five spaces for the As and one of the remaining three spaces for the E. Thus, the number of admissible arrangements is $$4!\binom{5}{2}\binom{3}{1} = 720$$ Method 2: We use the Inclusion-Exclusion Principle. There are seven letters in total, including two As, one B, one E, one G, one L, and one R. There are $\binom{7}{2}$ ways to choose two of the seven positions in the arrangement for the As. The remaining five letters can be placed in the remaining five positions in $5!$ ways. Hence, there are a total of $$\binom{7}{2}5! = \frac{7!}{2!5!} \cdot 5! = \frac{7!}{2!}$$ ways of arranging the letters of the word ALGEBRA. From these, we must subtract those arrangements in which one or more pairs of vowels are consecutive. A pair of vowels is consecutive: Two As are consecutive or an A and E are consecutive. Two As are consecutive: We have six objects to arrange. They are AA, B, E, G, L, R. Since they are distinct, they can be arranged in $6!$ ways. An A and an E are consecutive: We have six objects to arrange. They are B, E, G, L, R, and a block containing an A and an E. The six objects are distinct, so they can be arranged in $6!$ ways. The A and E can be arranged within the block in $2!$ ways. Hence, there are $6!2!$ such arrangements. If we subtract the number of arrangements in which a pair of vowels is consecutive from the total, we will have subtracted each arrangement in which two pairs of vowels are consecutive twice, once for each way we could have designated one of those pairs as the pair of consecutive vowels. Since we only want to subtract such arrangements once, we must add them to the total. Two pairs of consecutive vowels: Since there are only three vowels, this can only occur if the three vowels are consecutive. We have five objects to arrange, B, G, L, R, and the block of three vowels. Since the objects are distinct, they can be arranged in $5!$ ways. The three vowels can be arranged in three ways: AAE, AEA, EAA. Hence, the number of such arrangements is $5! \cdot 3$. By the Inclusion-Exclusion Principle, the number of admissible arrangements is $$\frac{7!}{2!} - 6! - 6!2! + 5! \cdot 3 = 720$$ The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is 151200. Explanation: Total number of words is INTERMEDIATE = 12 Which have 6 vowels and 6 consonants If two vowels never come together then we can arrange as under V C V C V C V C V C V C V Here, vowels are IEEIAE where 2 I’s and 3 E’s are there. ∴ Number of ways of arranging vowels = `(7!)/(3!2!)` = 420. Consonants are NTRMDT where 2T’s are there ∴ Number of ways arranging consonants = `(6!)/(2!)` = `(6*5*4*3*2!)/(21)` = 360 So, the total number of words are = 420 × 360 = 151200 Hence, the value of the filler is 151200.
Exercise :: Permutation and Combination - General Questions
Page 2
Exercise :: Permutation and Combination - General Questions
Improve Article Save Article Like Article Improve Article Save Article Given a word containing vowels and consonants. The task is to find that in how many ways the word can be arranged so that the vowels always come together. Given that the length of the word <10. Examples:
Approach: Since word contains vowels and consonants together. All vowels are needed to remain together then we will take all vowels as a single letter.
Below is the implementation of the above approach:
Time complexity: O(n) where n is the length of the string Further Optimizations: We can pre-compute required factorial values to avoid re-computations. |
The number of ways of arranging all the words of MAHESH such that all the vowels come together is
Pos Terkait
Copyright © 2024 ketajaman Inc.
