Your answer $7^4 \cdot 5^3$ is the number of ways the first four positions can be filled with consonants and the last three positions can be filled with vowels when letters may be repeated. However, the problem does not specify the positions of the consonants and vowels, so we must select them. Notice that selecting which four of the seven positions will be filled with consonants also determines the positions of the three vowels since each of the remaining three positions must be filled with vowels. We must select which four of the seven positions must be filled with consonants, which can be done in $\binom{7}{4}$ ways. Each of those four selected positions can be filled with a consonant in seven ways. Each of the remaining three positions can be filled with a vowel in five ways. Hence, there are $$\binom{7}{4}7^4 \cdot 5^3$$ words that can be formed consisting of four consonants and three vowels selected from an alphabet of seven consonants and three vowels with repetition. The stated answer $7^7\dbinom{7}{4}\dbinom{5}{3}$ is nonsense.
Your answer $P(7, 4)P(5, 3)$ is the number of ways of filling the first four positions with distinct consonants and the last three positions with distinct vowels. However, the problem does not specify the positions of the consonants and vowels, so we must select them. As above, we select four of the seven positions for the consonants. There are $P(7, 4)$ ways to arrange four of the seven consonants in those positions and $P(5, 3)$ ways to arrange three of the five vowels in the remaining three positions. Hence, there are $$\binom{7}{4}P(7, 4)P(5, 3)$$ words that can be formed consisting of four distinct consonants and three distinct vowels selected from an alphabet of seven consonants and five vowels. Notice that $$\binom{7}{4}P(7, 4)P(5, 3) = \frac{7!}{4!3!} \cdot \frac{7!}{3!} \cdot \frac{5!}{2!} = 7! \cdot \frac{7!}{4!3!} \cdot \frac{5!}{2!3!} = 7!\binom{7}{4}\binom{5}{3}$$ The author of your book arrived at the answer by selecting four of the seven consonants in $\binom{7}{4}$ ways, selecting three of the five vowels in $\binom{5}{3}$ ways, and then arranging the seven distinct selected letters in $7!$ ways.
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How many vowels can there be in a 5As mentioned above, there are no 5-letter words that consist of only vowels. However, there are 18 words that consist of 4 unique vowels. These words include: adieu, aequi, aoife, audio, aueto, auloi, aurei, avoue, heiau, kioea, louie, miaou, ouabe, ouija, oukia, ourie, ousia and uraei.
Is there a 5susus : plural of "susu", a money sharing group.
Does a five letter word need two vowels?There are many five letter words with two vowels. Apple, Heart, Pizza, Water, Music, Green and Death are examples.
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