Permutations & Combinations
Grade 11
Question:
<p>In how many ways can the letters of the word <strong>INTERMEDIATE</strong> be arranged so that the two vowels do not occur together?</p>
<p style="display:inline">151200</p>
<p style="display:inline">51200</p>
<p style="display:inline">15120</p>
<p style="display:inline">5040</p>
Step-by-Step Solution
Key Concept: The Gap Method is used to ensure no two vowels are adjacent by first arranging the consonants and then placing the vowels into the resulting spaces between them.
<p>There are 6 consonants (IN, IR, ID, IM, and 2T’s) and 6 vowels (2I’s, 3E’s, and 1A)<br />
Number of ways to arrange 6 consonants = <span class="math-tex">$\frac{6 !}{2 !}$</span><br />
Now, there are 7 gaps (available to arrange 6 vowels) created by these 6 consonants<br />
Number of ways to arrange 6 vowels in these 7 gaps = <span class="math-tex">$\frac{{ }^{7} \mathrm{P}_{6}}{2 ! 3 !}$</span><br />
<span class="math-tex">$\Rightarrow$</span> The required number of arrangements<br />
<span class="math-tex">$=\frac{6 !}{2 !} \times \frac{{ }^{7} \mathrm{P}_{6}}{2 ! 3 !}$</span> = 360 <span class="math-tex">$\times$</span> 420 = 151200</p>
Correct Answer: A