How many ways can the letters of the word committee be arranged so that the four vowels do not join together when read all at once?

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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.

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12 Questions 36 Marks 20 Mins

Calculations:

Number of alphabets in word TODAY = 5

Number of arrangements of word TODAY = 5!

⇒ Number of arrangements of word TODAY = 5 × 4 × 3 × 2 × 1

⇒ Number of arrangements of word TODAY = 120      ----(1)

Arrangement with vowels together = 4! = 24

Number of arrangements of vowels amongst themselves = 2! = 2

Number of arrangements with vowels together = 24 × 2 = 48      ----(2)

Required arrangements = Total arrangement - Arrangement with vowels together

⇒ Required number of arrangements = 120 - 48     [From (1) and (2)]

⇒ Required number of arrangements = 72

∴ The required number of arrangements is 72.

The number of ways of arranging unlike letters of an 'n' lettered word = n!

n! = n × (n - 1) × (n - 2) ×........× 1

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