Solveeit Logo
Login

Question

Question: How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the le...

How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word: DAUGHTER?

Explanation

Solution

Since we have to choose two vowels and three consonants out of the given word, we will apply the combination formula that is nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} , as we can find the ways in which we can select the required vowels and consonants. After finding the ways we will proceed with finding out the ways to arrange the selected letters, this we will do by applying the permutation formula that is nPr=n!(n−r)!^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}. Finally we will multiply both and get our answer.

Complete step by step answer:
We are given the word: DAUGHTER, in it we have a total of:
Vowels: A, U, E
Consonants: D, G, H, T, R
As we saw above that we have a total of 33 vowels and we have to choose 22 out of them. Therefore, the number of ways that we can choose the vowels will be: 3C2^{3}{{C}_{2}}
We know that nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} therefore:
3C2=3!2!(3−2)!=3!2!1!=3^{3}{{C}_{2}}=\dfrac{3!}{2!\left( 3-2 \right)!}=\dfrac{3!}{2!1!}=3
Similarly, We have a total of 5 consonants in the given word and we have to choose 3 out of those 5 consonants and hence the number of ways in which we can choose the consonants are: 5C3^{5}{{C}_{3}}
5C3=5!3!(5−3)!=5!3!2!=10^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!}=\dfrac{5!}{3!2!}=10
Now, we will find out the total number of ways in which we can select 22 vowels and 33 consonants: 3C2× 5C3=3×10=30^{3}{{C}_{2}}\times {{~}^{5}}{{C}_{3}}=3\times 10=30
Hence, the total number of ways of selecting 22 vowels and 33 consonants are 30. Now these 5 letters can be arranged in 5 different ways that means in 5P5^{5}{{P}_{5}} ways.
Now we know that nPr=n!(n−r)!^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}
Therefore: 5P5=5!(5−5)!=5!0!=5!1=5×4×3×2×1=120^{5}{{P}_{5}}=\dfrac{5!}{\left( 5-5 \right)!}=\dfrac{5!}{0!}=\dfrac{5!}{1}=5\times 4\times 3\times 2\times 1=120

Hence, total number of words will be: 30×120=360030\times 120=3600.

Note: You might think of doing this question in a manual way but as you saw the number that came is very large it is not possible to write down each combination and then arrange them accordingly. So it is better to proceed with the conventional formulas for permutation and combination. Be careful while taking out the factorials of the number.