How many arrangements can be made by using the letters of the word GARDEN so that the vowels are in alphabetical order?
Answer
602.4k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

