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## Kinh Nghiệm về How many 4 letter words can be formed using letters of word chemistry such that they start and end with vowel? Mới Nhất

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(Last Updated On: March 11, 2022)

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• Latest Problem Solving in Venn Diagram, Permutation, Combination and Probability
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• Find the number of four letter words that can be formed from the letters of the word HISTORY. (each letter to be used most once)(i)How many of them contain only constants?(ii)How many of them begin and end in a constant?(iii)How many of them begin with a vowel?(iv)How many contain the letter Y ?(v)How many begin with T and end in a vowel?(vi)How many begin with T and also contain S?(vii)How many contain both vowels?
• How many 4 letter words can be formed using the letters of the word chemistry such that they start and end with a vowel?
• How many 4 lettered words can be formed using the letters of the word chemistry such that they include all the vowels?
• How many 4 letter words can be formed from the letters of the word answer ‘? How many of these words start with a vowel?
• How many 4 letters words can be formed?

#### Problem Statement:

How many four-letter words beginning and ending with a vowel without any letter repeated can be formed from the word “personnel”?

There are 40 four-letter words beginning and ending with a vowel without any letter repeated can be formed.

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## Find the number of four letter words that can be formed from the letters of the word HISTORY. (each letter to be used most once)(i)How many of them contain only constants?(ii)How many of them begin and end in a constant?(iii)How many of them begin with a vowel?(iv)How many contain the letter Y ?(v)How many begin with T and end in a vowel?(vi)How many begin with T and also contain S?(vii)How many contain both vowels?

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Hint: We should use permutation and combination formulas to find desired results. We can use [^nC_rrm ,rm ^nP_r] to get results. [^nC_r] Is used whenever we need to do a combination of 1 or more elements.

Formula Used: [^nC_r = dfracn!r!left( n – r right)!rm ,rm ^nP_r = dfracn!left( n – r right)!]

Here each letter can only be used most one-time
which means there is no repetition in this.

(i)Part No 1st
So, we can use [^nP_r] to find the desired result.
Here n=total no. of letters in HISTORY word=7 and r=total no. of letters used to form new word=4
Putting these values in the formula [^nP_r = dfracn!left( n – r right)!] .
[Rightarrow ^7P_4 = dfrac7!left( 7 – 4 right)!]
Solving the factorials.
[ Rightarrow ^7P_4 = dfrac7*6*5*4*3!3!]
Cancelling 3! of
numerator From 3! In the denominator.
[ Rightarrow ^7P_4 = 840]
Which means 840 four letter words can be formed from the letters of the word HISTORY.

(ii)Part No 2nd
There are 5 constants in word HISTORY.
So n=5 and r=total no. of letters used to form new word=4
Putting these values in the formula [^nP_r = dfracn!left( n – r right)!] .
[Rightarrow ^5P_4 = dfrac5!left( 5 – 4 right)!]
Solving the factorials
[
Rightarrow ^5P_4 = 5! = 120]
Which means 120 four letter words can be formed which only contains constants.

(iii)Part No 3rd
There are 5 constants in word HISTORY out of which we will use 2 for begin and end.
So n=5 and r = begin + ending =2 .
Putting these values in the formula [^nP_r = dfracn!left( n – r right)!] .
[Rightarrow ^5P_2 = dfrac5!left( 5 – 2 right)!]
As there is no repetition, so only 5 words are left to fill 2
middle places.
So n=5 and r = 2 middle places.
Putting these values in the formula.
[Rightarrow ^5P_2 = dfrac5!left( 5 – 2 right)!]
Now, Multiplying All.
[ Rightarrow ^5P_2*^5P_2]
Solving the Formulas and factorials
[ Rightarrow ^5P_2 = 5*4 = 20]
[ Rightarrow ^5P_2*^5P_2 = 20*20 = 400]
Which means 400 four letter words can be formed which contain constants as their end and beginning letters.

(iv)Part No 4th

There are 2 vowels in word HISTORY out of which we will use 1 for beginning.
So n=2 and r = 1 .
Putting these values in the formula .
[Rightarrow ^2P_1 = dfrac2!left( 2 – 1 right)! = 2]
As there is no repetition, so only 6 words are left to fill 3 remaining places.
So n=6 and r = 3.
Putting these values in the formula.
[Rightarrow ^6P_3 = dfrac6!left( 6 – 3 right)! = 120]
Now, Multiplying All.
[ Rightarrow ^2P_1*^6P_3]
[
Rightarrow ^2P_1*^6P_3 = 2*120 = 240]
Which means 240 four letter words can be formed which contain vowels as their beginning letters.

(v)Part No 5th
There is only one Y in word HISTORY out of which we will use 1 for filling a place.
So, we will have to use a combination to find where to place Y.
So n=4 and r = 1 as Y is only one.
Putting these values in the formula [^nC_r = dfracn!r!left( n – r right)!] .
[^4C_1 = dfrac4!1!left(
4 – 1 right)! = 4]
Now we have 6 letters to fill 3 remaining places.
So n=6 and r = 3 .
Putting these values in the formula.
[Rightarrow ^6P_3 = dfrac6!left( 6 – 3 right)! = 120]
Now, Multiplying All.
[ Rightarrow ^4C_1*^6P_3]
[ Rightarrow ^4C_1*^6P_3 = 4*120 = 480]
Which means 480 four letter words can be formed which contains Y letter in it.

(vi)Part No 6th
There are 2 vowels in word HISTORY out of which we will
use 1 for ending.
So n=2 and r = 1 .
Putting these values in the formula .
[Rightarrow ^2P_1 = dfrac2!left( 2 – 1 right)! = 2]
As T is fixed in the Start we will count it as [Rightarrow ^1P_1 = 1]
As there is no repetition, so only 5 words are left to fill 2 remaining places.
So n=5 and r = 2.
Putting these values in the formula.
[Rightarrow ^5P_2 = dfrac5!left( 5 – 2 right)! = 20]
Now, Multiplying All.
[
Rightarrow ^2P_1*^5P_2*^1P_1]
[ Rightarrow ^2P_1*^5P_2*^1P_1 = 2*20*1 = 40]
Which means 40 four letter words can be formed which contain T as there beginning letter and vowels as there ending Letter.

(vii)Part No 7th
As T is fixed in the Start we will count it as [Rightarrow ^1P_1 = 1]
We have 3 locations to choose from to place S. So, we will use Combination here.
[Rightarrow ^3C_1 = dfrac3!1!left( 3 – 1 right)! =
3]
As there is no repetition, so only 5 words are left to fill 2 remaining places.
So n=5 and r = 2.
Putting these values in the formula.
[Rightarrow ^5P_2 = dfrac5!left( 5 – 2 right)! = 20]
Now, Multiplying All.
[ Rightarrow ^3C_1*^5P_2*^1P_1]
[ Rightarrow ^3C_1*^5P_2*^1P_1 = 3*20*1 = 60]
Which means 60 four letter words can be formed which contain T as their beginning letter and S in any remaining place.

(viii)Part
No 8th
We have 4 locations to choose from to place both vowels. So, we will use Combination here.
[Rightarrow ^4C_2 = dfrac4!2!left( 4 – 2 right)! = 6]
As we are keeping both vowels we will also check their positions via permutation.
[Rightarrow ^2P_2 = dfrac2!left( 2 – 2 right)! = 2! = 2]
As there is no repetition, so only 5 words are left to fill 2 remaining places.
So n=5 and r = 2.
Putting these values in the formula.
[Rightarrow
^5P_2 = dfrac5!left( 5 – 2 right)! = 20]
Now, Multiplying All.
[ Rightarrow ^4C_2*^2P_2*^5P_2]
[ Rightarrow ^4C_2*^2P_2*^5P_2 = 6*2*20 = 240]
Which means 240 four letter words can be formed which contain both vowels in it.

Note: We should always take care of where we should use permutation and where combination both seems confusing but have defined work. Combinations are used for groups here order doesn’t matter
and Permutations are used for creating multiple lists.

### How many 4 letter words can be formed using the letters of the word chemistry such that they start and end with a vowel?

`therefore` The number of 4 letter words that begin and end with vowels `=6xx20=120` ways. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

### How many 4 lettered words can be formed using the letters of the word chemistry such that they include all the vowels?

Now, Multiplying All. Which means 240 four letter words can be formed which contain both vowels in it.

### How many 4 letter words can be formed from the letters of the word answer ‘? How many of these words start with a vowel?

Hence, 360 ways of 4 letter words can be formed from the letters of the word ‘ANSWER’ and 120 ways of 4 letter words start with vowels.

### How many 4 letters words can be formed?

Hence, The total number of four-letter words that can be formed is 270.
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