There are 3 consonants and 3 vowels.
So, 4 consonants and 2 vowels can be selected in 8C4 × 3C2 ways.
Now, 8C4 × 3C2
= `[[8 xx 7 xx 6 xx 5]]/[[4 xx 3 xx 2]] xx [[3 xx 2 xx 1]]/[[2 xx 1]`
= 70 × 3
= 210
Thus, there are 210 groups consisting of 4 consonants and 2 vowels.
We need to form different words from these 210 groups.
Now, each group has 6 letters.
These 6 letters can be arranged amongst themselves m 6! Ways.
∴ The number of required words
= [210] × 6!
= [210] × 720
= 151200
Note: In this question, students must take care that here the vowels as well as the consonants are repeating in nature as the word ‘DIFFERENTIATION” has more than one similar vowels and consonants and so we need to apply the arrangement in that also.
So I am having some trouble with a few parts of my homework problem. The question gives us 3 vowels [A,E,O] and 4 consonants [B,C,D,F].
a] How many ways can you make a 7 letter word if each letter can only be used once. [Word doesn't have to be real]. This was easy, as the answer is just 7!
b] If the vowels have to be together and the consonants have to be together?
My approach: 3 vowels can be rearranged in 3! ways and 4 consonants in 4! ways, and the vowels can be leading or ending, so the total is 3! * 4! * 2!.
c] If the vowels have to be together?
My approach: 3 vowels in 3! ways, and can be rearranged in 5! ways.
d] If B and C have to be together, but no other vowels or consonants can be together?
My approach: I first tried making a word that starts with a consonant, and then alternating between vowel and consonant, while keeping B and C last. My other word started with a vowel then a consonant, then B and C, then vowel, consonant and the vowel. This is all I could think of, and I'm not sure how the math checks out on this.
Solution[By Examveda Team]
4 consonants out of 12 can be selected in,12C4 ways.
3 vowels can be selected in 4C3 ways.
Therefore, total number of groups each containing 4 consonants and 3 vowels,
= 12C4 × 4C3 Each group contains 7 letters, which can be arranging in 7! ways.
Therefore required number of words,
= 12C4 × 4C3 × 7!