Answer VerifiedHint- In this question we use the theory of permutation and combination. We have to choose out of 7 Consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed. For example, in this case, how we choose two vowels out of four vowels and this can be done in ${}^{\text{4}}{{\text{C}}_2}$ =6 ways.Complete step-by-step answer:As we know,${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) is calculated using the formula-${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$Now,${}^{\text{7}}{{\text{C}}_{\text{3}}} \times {}^{\text{4}}{{\text{C}}_{\text{2}}}$= $\dfrac{{{\text{7!}}}}{{{\text{(7 - 3)!3!}}}} \times \dfrac{{{\text{4!}}}}{{{\text{(4 - 2)!2!}}}}$ = $\dfrac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \dfrac{{4 \times 3}}{{2 \times 1}}$ = 210Number of groups, each having 3 consonants and 2 vowels =210.Each group contains 5 letters.Number of ways of arranging 5 letters among themselves =5! = 5×4×3×2×1 = 120∴ Required number of ways = (210×120) =25200.Hence, the answer is 25200.So, option (A) is the correct answer.Note- Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. Thus, if we want to figure out how many combinations we have of n objects then r at a time, we just create all the permutations and then divide by r! variant. A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen? 4 consonants can be selected from 8 consonants in 8C4 ways and 2 vowels can be selected from 3 vowels in 3C2 ways. ∴ the number of words with 4 consonants and 2 vowels = 8C4 × 3C2 = `(8!)/(4!4!) xx (3!)/(2!1!)` = `(8 xx 7 xx 6 xx 5)/(4 xx 3 xx 2 xx 1) xx (3 xx 2!)/(2!)` = 70 × 3 = 210 Now each of these words contains 6 letters which can be arranged in 6P6 = 6! ways. ∴ the total number of words that can be formed with 4 consonants and 2 vowels = 210 × 6! = 210 × 6 × 5 × 4 × 3 × 2 × 1 = 151200. Concept: Properties of Combinations Is there an error in this question or solution? Open in App Suggest Corrections 0
Discussion :: Permutation and Combination - General Questions (Q.No.4)
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