Answer
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Hint: We will look at the number of vowels in the word GARDEN and their alphabetical order. Then we will count the number of ways to arrange these vowels so that their alphabetical order is maintained. We will look at the remaining letters, which are the consonants of the word GARDEN. We will have to count the number of ways these letters can be arranged. Then, we will find the total number of possible arrangements of the letters of the word GARDEN which satisfy the given conditions. Here we arrange n elements at n places by $n!= n\times(n-1)\times (n-2).......3\times 2 \times 1$
Complete step-by-step solution:
There are 6 letters in the word GARDEN. Two of them are vowels, A and E. These vowels have to be in alphabetical order, which means A has to be placed before E in every arrangement. Let us look at the arrangement as putting the letters into six slots. If we fix the letter A in slot 1, there are 5 ways to assign a position to letter E. If we fix the position of letter A in slot 2, the letter E can be placed in 4 different ways. Continuing in this manner, we can see that letter A in position 3 leaves us with 3 ways to place letter E. Letter A in position 4 gives us 2 more ways to place letter E, and the letter A in the fifth position gives us only one choice to put the letter E in slot 6. The letter A cannot be placed in slot 6 since there will be no way left to place the letter E so that the alphabetical order is maintained.
So, the number of ways to arrange the vowels of the letter GARDEN in alphabetical order is
$5+4+3+2+1=15$
Now, the remaining letter of the word GARDEN is the 4 consonants. Taking into account two slots for the vowels, these four consonants can be arranged in the remaining four slots in $4!$ ways.
Therefore, the total number of arrangements of all the letters of the word GARDEN such that the vowels are in alphabetical order is $15\times 4!=15\times 24=360$.
Note: It is important to understand the conditions asked for in the question, it will be useful to apply restrictions to the arrangements we count. While solving such types of questions, we should know when to add the count and when to multiply it. It is essential that we write down each case explicitly so that we can avoid any minor mistakes while counting.
Complete step-by-step solution:
There are 6 letters in the word GARDEN. Two of them are vowels, A and E. These vowels have to be in alphabetical order, which means A has to be placed before E in every arrangement. Let us look at the arrangement as putting the letters into six slots. If we fix the letter A in slot 1, there are 5 ways to assign a position to letter E. If we fix the position of letter A in slot 2, the letter E can be placed in 4 different ways. Continuing in this manner, we can see that letter A in position 3 leaves us with 3 ways to place letter E. Letter A in position 4 gives us 2 more ways to place letter E, and the letter A in the fifth position gives us only one choice to put the letter E in slot 6. The letter A cannot be placed in slot 6 since there will be no way left to place the letter E so that the alphabetical order is maintained.
So, the number of ways to arrange the vowels of the letter GARDEN in alphabetical order is
$5+4+3+2+1=15$
Now, the remaining letter of the word GARDEN is the 4 consonants. Taking into account two slots for the vowels, these four consonants can be arranged in the remaining four slots in $4!$ ways.
Therefore, the total number of arrangements of all the letters of the word GARDEN such that the vowels are in alphabetical order is $15\times 4!=15\times 24=360$.
Note: It is important to understand the conditions asked for in the question, it will be useful to apply restrictions to the arrangements we count. While solving such types of questions, we should know when to add the count and when to multiply it. It is essential that we write down each case explicitly so that we can avoid any minor mistakes while counting.
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