The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is :
Solution
In the given word, <br/><br/>vowels are : I, E, E, E, E <br/><br/>Consonants are : N, D, P, N, D, N, C <br/><br/>So, number of words $=\frac{8 !}{3 ! 2 !} \times \frac{5 !}{4 !}$ <br/><br/>$=\frac{8 \times 7 \times 6 \times 5 \times 4}{2} \times 5=16800$
<br/><br/><b>Concept :</b>
<br/><br/>Out of $n$ objects, if $r$ things are same, so number of ways $=\frac{n !}{r !}$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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