Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Suggest Corrections 3 Answer & Explanation 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 = $$\frac{8 !}{(2 !) (2 !)}$$ = 10080. Now, AEAI has 4 Letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = $$\frac{4 !}{2 !}$$ = 12. $$\therefore$$ Required number of words = (10080 * 12) = 120960.
Solution: The word ‘STRANGE’ has seven letters, including two vowels (A,E) and five consonants (S,T,R,N,G). (i) the vowels come together? If we consider two vowels to be one letter, we’ll end up with six letters that can be ordered in six different ways. (A,E) can be combined in 2P2 ways. As a result, the needed word count is Using the formula, we can $ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! $ $ P\text{ }\left( 6,\text{ }6 \right)\text{ }\times \text{ }P\text{ }\left( 2,\text{ }2 \right)\text{ }=\text{ }6!/\left( 6-6 \right)!\text{ }\times \text{ }2!/\left( 2-2 \right)! $ $ =\text{ }6!\text{ }\times \text{ }2! $ $ =\text{ }6\text{ }\times \text{ }5\text{ }\times \text{ }4\text{ }\times \text{ }3\text{ }\times \text{ }2\text{ }\times \text{ }1\text{ }\times \text{ }2\text{ }\times \text{ }1 $ $ =\text{ }720\text{ }\times \text{ }2 $ $ =\text{ }1440 $ (ii) the vowels never come together? The total number of letters in the word ‘STRANGE’ is given by: 7P7 = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 7P7 = 5040 So, Total number of words where vowels aren’t together = total number of words – total number of words in which vowels are always together = 5040 – 1440 = 3600 As a result, there are 3600 configurations in which vowels never come together.
Exercise :: Permutation and Combination - General Questions
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Exercise :: Permutation and Combination - General Questions
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