To solve this problem, we need to consider the word "EASYQUIZ" and the condition that the vowels always come together.

Step 1: Identify the vowels and consonants in the word. The word "EASYQUIZ" has 4 vowels (E, A, U, I) and 4 consonants (S, Y, Q, Z).

Step 2: Since the vowels need to always come together, we can consider them as a single letter. So, we have 5 "letters" - the 4 consonants and the "group" of vowels.

Step 3: Calculate the number of ways these 5 "letters" can be arranged. This is a permutation problem, so we use the formula for permutations of n items, which is n!. So, the number of ways to arrange these 5 "letters" is 5! = 120.

Step 4: However, we also need to consider the different ways the vowels within the "group" can be arranged. This is also a permutation problem, so we use the same formula. The number of ways to arrange the 4 vowels is 4! = 24.

Step 5: Multiply the two results together to get the total number of possible words. So, the total number of possible words is 120 * 24 = 2880.

So, there are 2880 possible words that can be made using the word “EASYQUIZ” such that the vowels always come together.

Question

To solve this problem, we need to consider the word "EASYQUIZ" and the condition that the vowels always come together.

Step 1: Identify the vowels and consonants in the word. The word "EASYQUIZ" has 4 vowels (E, A, U, I) and 4 consonants (S, Y, Q, Z).

Step 2: Since the vowels need to always come together, we can consider them as a single letter. So, we have 5 "letters" - the 4 consonants and the "group" of vowels.

Step 3: Calculate the number of ways these 5 "letters" can be arranged. This is a permutation problem, so we use the formula for permutations of n items, which is n!. So, the number of ways to arrange these 5 "letters" is 5! = 120.

Step 4: However, we also need to consider the different ways the vowels within the "group" can be arranged. This is also a permutation problem, so we use the same formula. The number of ways to arrange the 4 vowels is 4! = 24.

Step 5: Multiply the two results together to get the total number of possible words. So, the total number of possible words is 120 * 24 = 2880.

So, there are 2880 possible words that can be made using the word “EASYQUIZ” such that the vowels always come together.

Knowee AI · Accepted Answer

To solve this problem, we need to consider the word "EASYQUIZ" and the condition that the vowels always come together.

Step 1: Identify the vowels and consonants in the word. The word "EASYQUIZ" has 4 vowels (E, A, U, I) and 4 consonants (S, Y, Q, Z).

Step 2: Since the vowels need to always come together, we can consider them as a single letter. So, we have 5 "letters" - the 4 consonants and the "group" of vowels.

Step 3: Calculate the number of ways these 5 "letters" can be arranged. This is a permutation problem, so we use the formula for permutations of n items, which is n!. So, the number of ways to arrange these 5 "letters" is 5! = 120.

Step 4: However, we also need to consider the different ways the vowels within the "group" can be arranged. This is also a permutation problem, so we use the same formula. The number of ways to arrange the 4 vowels is 4! = 24.

Step 5: Multiply the two results together to get the total number of possible words. So, the total number of possible words is 120 * 24 = 2880.

So, there are 2880 possible words that can be made using the word “EASYQUIZ” such that the vowels always come together.

What is the number of possible words that can be made using the word “EASYQUIZ” such that the vowels always come together?

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