How many words can be formed from the letters of the word engineering so that vowels always come together?

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Exercise :: Permutation and Combination - General Questions

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2.

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A. 360 B. 480 C. 720 D. 5040 E. None of these Answer: Option C Explanation: The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720. Video Explanation: https://youtu.be/WCEF3iW3H2c

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Exercise :: Permutation and Combination - General Questions

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7.

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Answer: Option D Explanation: Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it. The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place. The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it. Required number of numbers = (1 x 5 x 4) = 20.

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Exercise :: Permutation and Combination - General Questions

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13.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A. 10080 B. 4989600 C. 120960 D. None of these Answer: Option C Explanation: In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. Number of ways of arranging these letters = 8! = 10080. (2!)(2!) Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4! = 12. 2! Required number of words = (10080 x 12) = 120960.