
Find the number of words that can be formed out of the letters $a,b,c,d,e,f$ taken $3$ together, each containing one vowel at least.
Answer
522k+ views
Hint: In this problem we need to find the number of $3$ letter words, each containing one vowel at least. For this, first, we will find a number of words each containing exactly one vowel. Then, we will find a number of words each containing exactly two vowels. In this problem, we have only two vowels $a$ and $e$. We will add the number of words containing exactly one vowel and the number of words containing exactly two vowels to get the total number of three-letter words. We will use the concept of combination and permutation in this problem.
Complete step by step answer:
We have two vowels $a,e$ and four consonants $b,c,d,f$. To find the number of three-letter words each containing one vowel at least, we will consider the following cases:
The case I: Exactly one vowel and two consonants.
In this case, one vowel out of two vowels can be selected in ${}^2{C_1} = 2$ ways. Two consonants out of four consonants can be selected in ${}^4{C_2} = \dfrac{{4 \times 3}}{{1 \times 2}} = \dfrac{{12}}{2} = 6$ ways.
All three letters (one vowel and two consonants) can arrange among themselves in $3! = 6$ ways. Therefore, the total number of three-letter words each containing exactly one vowel are $2 \times 6 \times 6 = 72$.
Case II: Exactly two vowels and one consonant.
Two vowels out of two vowels can be selected in ${}^2{C_2} = \dfrac{{2 \times 1}}{{1 \times 2}} = 1$ way. One consonant out of four consonants can be selected in ${}^4{C_1} = 4$ ways.
All three letters (one vowel and two consonants) can arrange among themselves in $3! = 6$ ways. Therefore, the total number of three-letter words each containing exactly two vowels is $1 \times 4 \times 6 = 24$. Now we will add the possibilities of the case I and case II to get the required number of words. Therefore, the total number of three-letter words each containing at least one vowel is $72 + 24 = 96$.
Note:
If we consider the order of objects then it is a permutation. If we do not consider the order of objects then it is a combination. ${}^n{C_r}$ gives the total number of ways of selecting $r$ objects out of $n$ objects. ${}^n{P_r}$ gives the total number of distinct arrangements when we arrange $r$ objects among $n$ objects. The number of permutations of $n$ distinct objects is $n!$.
Complete step by step answer:
We have two vowels $a,e$ and four consonants $b,c,d,f$. To find the number of three-letter words each containing one vowel at least, we will consider the following cases:
The case I: Exactly one vowel and two consonants.
In this case, one vowel out of two vowels can be selected in ${}^2{C_1} = 2$ ways. Two consonants out of four consonants can be selected in ${}^4{C_2} = \dfrac{{4 \times 3}}{{1 \times 2}} = \dfrac{{12}}{2} = 6$ ways.
All three letters (one vowel and two consonants) can arrange among themselves in $3! = 6$ ways. Therefore, the total number of three-letter words each containing exactly one vowel are $2 \times 6 \times 6 = 72$.
Case II: Exactly two vowels and one consonant.
Two vowels out of two vowels can be selected in ${}^2{C_2} = \dfrac{{2 \times 1}}{{1 \times 2}} = 1$ way. One consonant out of four consonants can be selected in ${}^4{C_1} = 4$ ways.
All three letters (one vowel and two consonants) can arrange among themselves in $3! = 6$ ways. Therefore, the total number of three-letter words each containing exactly two vowels is $1 \times 4 \times 6 = 24$. Now we will add the possibilities of the case I and case II to get the required number of words. Therefore, the total number of three-letter words each containing at least one vowel is $72 + 24 = 96$.
Note:
If we consider the order of objects then it is a permutation. If we do not consider the order of objects then it is a combination. ${}^n{C_r}$ gives the total number of ways of selecting $r$ objects out of $n$ objects. ${}^n{P_r}$ gives the total number of distinct arrangements when we arrange $r$ objects among $n$ objects. The number of permutations of $n$ distinct objects is $n!$.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Write examples of herbivores carnivores and omnivo class 10 biology CBSE

Give 10 examples of Material nouns Abstract nouns Common class 10 english CBSE

Difference between mass and weight class 10 physics CBSE
