Find the number of four letter words that can be formed from the letters of the word HISTORY. (each letter to be used at 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?
Answer
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Hint: We should use permutation and combination formulas to find desired results. We can use \[{}^n{C_r}{\rm{ }},{\rm{ }}{}^n{P_r}\] to get results. \[{}^n{C_r}\] Is used whenever we need to do a combination of 1 or more elements.
Formula Used: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}{\rm{ }},{\rm{ }}{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Complete step-by-step answer:
Here each letter can only be used at most one-time which means there is no repetition in this.
(i)Part No 1st
So, we can use \[{}^n{P_r}\] 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 \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] .
\[\Rightarrow {}^7{P_4} = \dfrac{{7!}}{{\left( {7 - 4} \right)!}}\]
Solving the factorials.
\[ \Rightarrow {}^7{P_4} = \dfrac{{7*6*5*4*3!}}{{3!}}\]
Cancelling 3! of numerator From 3! In the denominator.
\[ \Rightarrow {}^7{P_4} = 840\]
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 \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] .
\[\Rightarrow {}^5{P_4} = \dfrac{{5!}}{{\left( {5 - 4} \right)!}}\]
Solving the factorials
\[ \Rightarrow {}^5{P_4} = 5! = 120\]
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 \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] .
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}}\]
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.
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}}\]
Now, Multiplying All.
\[ \Rightarrow {}^5{P_2}*{}^5{P_2}\]
Solving the Formulas and factorials
\[ \Rightarrow {}^5{P_2} = 5*4 = 20\]
\[ \Rightarrow {}^5{P_2}*{}^5{P_2} = 20*20 = 400\]
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 .
\[\Rightarrow {}^2{P_1} = \dfrac{{2!}}{{\left( {2 - 1} \right)!}} = 2\]
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.
\[\Rightarrow {}^6{P_3} = \dfrac{{6!}}{{\left( {6 - 3} \right)!}} = 120\]
Now, Multiplying All.
\[ \Rightarrow {}^2{P_1}*{}^6{P_3}\]
\[ \Rightarrow {}^2{P_1}*{}^6{P_3} = 2*120 = 240\]
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 \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] .
\[{}^4{C_1} = \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} = 4\]
Now we have 6 letters to fill 3 remaining places.
So n=6 and r = 3 .
Putting these values in the formula.
\[\Rightarrow {}^6{P_3} = \dfrac{{6!}}{{\left( {6 - 3} \right)!}} = 120\]
Now, Multiplying All.
\[ \Rightarrow {}^4{C_1}*{}^6{P_3}\]
\[ \Rightarrow {}^4{C_1}*{}^6{P_3} = 4*120 = 480\]
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 .
\[\Rightarrow {}^2{P_1} = \dfrac{{2!}}{{\left( {2 - 1} \right)!}} = 2\]
As T is fixed in the Start we will count it as \[\Rightarrow {}^1{P_1} = 1\]
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.
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}} = 20\]
Now, Multiplying All.
\[ \Rightarrow {}^2{P_1}*{}^5{P_2}*{}^1{P_1}\]
\[ \Rightarrow {}^2{P_1}*{}^5{P_2}*{}^1{P_1} = 2*20*1 = 40\]
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 \[\Rightarrow {}^1{P_1} = 1\]
We have 3 locations to choose from to place S. So, we will use Combination here.
\[\Rightarrow {}^3{C_1} = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} = 3\]
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.
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}} = 20\]
Now, Multiplying All.
\[ \Rightarrow {}^3{C_1}*{}^5{P_2}*{}^1{P_1}\]
\[ \Rightarrow {}^3{C_1}*{}^5{P_2}*{}^1{P_1} = 3*20*1 = 60\]
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.
\[\Rightarrow {}^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} = 6\]
As we are keeping both vowels we will also check their positions via permutation.
\[\Rightarrow {}^2{P_2} = \dfrac{{2!}}{{\left( {2 - 2} \right)!}} = 2! = 2\]
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.
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}} = 20\]
Now, Multiplying All.
\[ \Rightarrow {}^4{C_2}*{}^2{P_2}*{}^5{P_2}\]
\[ \Rightarrow {}^4{C_2}*{}^2{P_2}*{}^5{P_2} = 6*2*20 = 240\]
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.
Formula Used: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}{\rm{ }},{\rm{ }}{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Complete step-by-step answer:
Here each letter can only be used at most one-time which means there is no repetition in this.
(i)Part No 1st
So, we can use \[{}^n{P_r}\] 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 \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] .
\[\Rightarrow {}^7{P_4} = \dfrac{{7!}}{{\left( {7 - 4} \right)!}}\]
Solving the factorials.
\[ \Rightarrow {}^7{P_4} = \dfrac{{7*6*5*4*3!}}{{3!}}\]
Cancelling 3! of numerator From 3! In the denominator.
\[ \Rightarrow {}^7{P_4} = 840\]
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 \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] .
\[\Rightarrow {}^5{P_4} = \dfrac{{5!}}{{\left( {5 - 4} \right)!}}\]
Solving the factorials
\[ \Rightarrow {}^5{P_4} = 5! = 120\]
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 \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] .
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}}\]
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.
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}}\]
Now, Multiplying All.
\[ \Rightarrow {}^5{P_2}*{}^5{P_2}\]
Solving the Formulas and factorials
\[ \Rightarrow {}^5{P_2} = 5*4 = 20\]
\[ \Rightarrow {}^5{P_2}*{}^5{P_2} = 20*20 = 400\]
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 .
\[\Rightarrow {}^2{P_1} = \dfrac{{2!}}{{\left( {2 - 1} \right)!}} = 2\]
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.
\[\Rightarrow {}^6{P_3} = \dfrac{{6!}}{{\left( {6 - 3} \right)!}} = 120\]
Now, Multiplying All.
\[ \Rightarrow {}^2{P_1}*{}^6{P_3}\]
\[ \Rightarrow {}^2{P_1}*{}^6{P_3} = 2*120 = 240\]
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 \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] .
\[{}^4{C_1} = \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} = 4\]
Now we have 6 letters to fill 3 remaining places.
So n=6 and r = 3 .
Putting these values in the formula.
\[\Rightarrow {}^6{P_3} = \dfrac{{6!}}{{\left( {6 - 3} \right)!}} = 120\]
Now, Multiplying All.
\[ \Rightarrow {}^4{C_1}*{}^6{P_3}\]
\[ \Rightarrow {}^4{C_1}*{}^6{P_3} = 4*120 = 480\]
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 .
\[\Rightarrow {}^2{P_1} = \dfrac{{2!}}{{\left( {2 - 1} \right)!}} = 2\]
As T is fixed in the Start we will count it as \[\Rightarrow {}^1{P_1} = 1\]
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.
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}} = 20\]
Now, Multiplying All.
\[ \Rightarrow {}^2{P_1}*{}^5{P_2}*{}^1{P_1}\]
\[ \Rightarrow {}^2{P_1}*{}^5{P_2}*{}^1{P_1} = 2*20*1 = 40\]
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 \[\Rightarrow {}^1{P_1} = 1\]
We have 3 locations to choose from to place S. So, we will use Combination here.
\[\Rightarrow {}^3{C_1} = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} = 3\]
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.
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}} = 20\]
Now, Multiplying All.
\[ \Rightarrow {}^3{C_1}*{}^5{P_2}*{}^1{P_1}\]
\[ \Rightarrow {}^3{C_1}*{}^5{P_2}*{}^1{P_1} = 3*20*1 = 60\]
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.
\[\Rightarrow {}^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} = 6\]
As we are keeping both vowels we will also check their positions via permutation.
\[\Rightarrow {}^2{P_2} = \dfrac{{2!}}{{\left( {2 - 2} \right)!}} = 2! = 2\]
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.
\[\Rightarrow {}^5{P_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!}} = 20\]
Now, Multiplying All.
\[ \Rightarrow {}^4{C_2}*{}^2{P_2}*{}^5{P_2}\]
\[ \Rightarrow {}^4{C_2}*{}^2{P_2}*{}^5{P_2} = 6*2*20 = 240\]
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.
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