Free Show
General Knowledge: Free Mock Test 10 Questions 10 Marks 7 Mins Given: 8 consonants and 4 vowels Calculation: According to the question 3 consonants out of 8 can be selected in 8C3 = 56 2 vowels out of 4 can be selected in 4C2 = 6 So, five letters of word can be chosen = 56 × 6 = 336 But all these five alphabets can permute among themselves = 5! The number of word = 336 × 5! = 40320 ∴ The required result will be 40320.
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. From $5$ consonants and $4$ vowels, how many words can be formed by using $3$ consonants and $2$ vowels.A. 9440 B. 6800 C. 3600D. 7200Answer Verified Hint: The number of ways a word can form from $5$ consonants by using $3$ consonants $ = $ ${}^5{C_3}$ and from $4$ vowels by using $2$ vowels $ = $${}^4{C_2}$, hence the number of words can be $ = {}^5{C_3} \times {}^4{C_2} \times {}^5{P_5}$. Use this to find the no. of words. Complete step-by-step solution: Note: It is advisable in such types of questions we should see that what are all possibilities that words can be formed , for this one must have a basic understanding of permutation and combination. Here we have used $ {}5{P_5}$ for arranging 5 words. How many words of 4 consonants and 4 vowels can be formed out of 8 vowels and 5 Consonents?Detailed Solution
But all these five alphabets can permute among themselves = 5! ∴ The required result will be 40320.
How many words can be formed by taking 3 consonants and 2 vowels out of 5 consonants and 4 vowels?From 5 consonants and 4 vowels, how many words can be formed by using 3 consonants and 2 vowels. A. 9440.
How many words of 2 consonants and 2 vowels can be formed from 5 consonants and 4 vowels?⇒ Number of ways selecting 2 consonants = (5 × 4 × 3!)/(2 × 1 × 3!) ∴ The total number of ways is 1440.
How many words can be formed consisting of 2 consonants and 2 vowels from 4 consonants and 3 vowels?= 210. Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters. Required number of ways = (210 x 120) = 25200.
...
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?. |