# CRE 2 - Using letters of a word | Modern Math - Permutation & Combination

**Answer the next 10 questions based on the information given:**

Consider the letters of the word “JOURNEY”. Using the letters of this word,

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

How many 7 letter words can be formed?

Answer: 5040

**Explanation** :

Number of ways of arranging 7 letters = 7! = 5040.

Hence, 5040.

Workspace:

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

How many 7 letter words starting with O can be formed?

Answer: 720

**Explanation** :

If the first position is occupied by O, we have 6 positions left which will be occupied by 6 letters.

∴ Number of words = 6! = 720.

Hence, 720.

Workspace:

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

How many 7 letter words ending with Y can be formed?

Answer: 720

**Explanation** :

If the last position is occupied by Y, we have 6 positions left which will be occupied by 6 letters.

∴ Number of words = 6! = 720.

Hence, 720.

Workspace:

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

How many 7 letter words starting with O & ending with Y can be formed?

Answer: 120

**Explanation** :

If the first position is occupied by O and last by Y, we have 5 positions left which will be occupied by 5 letters.

∴ Number of words = 5! = 120.

Hence, 120.

Workspace:

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

How many 7 letter words starting with O or ending with Y can be formed?

Answer: 120

**Explanation** :

Number of words starting with O or ending with Y = O + Y – O ∩ Y.

O = Number of words starting with O = 720.

Y = Number of words ending with Y = 720.

O ∩ Y = Number of words starting with O & ending with Y = 120.

∴ O + Y – O ∩ Y = 720 + 720 – 120 = 1320.

Hence, 1320.

Workspace:

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

How many 7 letter words neither starting with O nor ending with Y can be formed?

Answer: 3720

**Explanation** :

Total number of words = 7! = 5040. (From 1^{st} question)

Number of words starting with O or ending with Y = 1320 (From previous question)

∴ Number of words neither starting with O or ending with Y = 5040 – 1320 = 3720.

Hence, 3720.

Workspace:

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

How many 7 letter words can be formed such that all the vowels are together?

Answer: 720

**Explanation** :

There are three vowel in JOURNEY i.e., E, O and U.

All three vowels are together. Let’s call the group of vowels as X.

⇒ We have 5 letters i.e., J R N Y X which can be arranged in 5! = 120 ways.

For each of these 120 arrangements, internal arrangement of vowels in X can be done in 3! = 6 ways.

∴ Number of words that can be formed such that all the vowels are together = 120 × 6 = 720 ways.

Hence, 720 ways.

Workspace:

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

How many 7 letter words can be formed such that no two vowels are together?

Answer: 1440

**Explanation** :

Let’s first arrange the remaining letters i.e., J, R, N, and Y

Number of ways of arranging these 4 letters = 4! = 24 ways.

Let’s take one of these arrangements as J R N Y.

Now we have 5 places to arrange the remaining 3 vowels i.e., | J | R | N | Y|.

Hence, 3 vowels can be arranged in 5 places in 5 × 4 × 3 = 60 ways.

∴ Total number of words that can be formed such that no two vowels are together = 24 × 60 = 1440.

Hence, 1440.

Workspace:

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

How many 7 letter words can be formed such that all the vowels are not together?

Answer: 4320

**Explanation** :

Total number of words = 7! = 5040. (From 1^{st} question)

Number of words where all the vowels are together = 720 ways.

∴ Number of words that can be formed such that all the vowels are not together = 5040 – 720 = 4320.

Hence, 4320.

Workspace:

**CRE 2 - Using letters of a word | Modern Math - Permutation & Combination**

What is the rank of the word “JOURNEY” among all the 7 letter words that can be formed using its letters in dictionary order?

Answer: 1047

**Explanation** :

Let’s first arrange all the letters in alphabetical order i.e., E J N O R U Y.

Let the position of letters be: 1 2 3 4 5 6 7

**Letter at 1**^{st}** place:**

Number of words starting with E = 6! = 720.

All these words will come before ‘JOURNEY’. From 721^{st} word onwards, first letter will be J.

J 2 3 4 5 6 7

**Letter at 2**^{nd}** place:**

Number of words with E at 2^{nd} place = 5! = 120.

Number of words with N at 2^{nd} place = 5! = 120.

Total 720 + 240 = 960 words.

All these words will come before ‘JOURNEY’. From 961^{st} word onwards, 1^{st} letter will be J and 2^{nd} letter will be O.

J O 3 4 5 6 7

**Letter at 3**^{rd}** place:**

Number of words with E at 3^{rd} place = 4! = 24.

Number of words with N at 3^{rd} place = 4! = 24.

Number of words with R at 3^{rd} place = 4! = 24.

Total 960 + 24 + 24 + 24 = 1032 words.

All these words will come before ‘JOURNEY’. From 1033^{rd} word onwards, 1st letter will be J, 2^{nd} letter will be O and 3^{rd} letter will be U.

J O U 4 5 6 7

**Letter at 4**^{th}** place:**

Number of words with E at 4^{th} place = 3! = 6.

Number of words with N at 4^{th} place = 3! = 6.

Total 1032 + 6 + 6 = 1044 words.

All these words will come before ‘JOURNEY’. From 1045^{th} word onwards, 1st letter will be J, 2^{nd} letter will be O, 3^{rd} letter will be U and 4^{th} letter will be R.

J O U R 5 6 7

**Letter at 5**^{th}** place:**

Number of words with E at 5^{th} place = 2! = 2.

Total 1044 + 2 = 1046 words.

All these words will come before ‘JOURNEY’. From 1047^{th} word onwards, 1^{st} letter will be J, 2^{nd} letter will be O, 3^{rd} letter will be U, 4^{th} letter will be R and 5^{th} letter will be N.

J O U R N 6 7

Now, 1047^{th} word will have E at 6^{th} and Y at 7^{th} position which makes the work J O U R N E Y.

∴ JOURNEY is ranked 1047^{th}.

Hence, 1047.

Workspace:

## Feedback

Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.