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. question total number of word formed by two vowels and 3 consonants taken from Four Women and 5% is equal to option 60 option B 120 option C 7200 Aarakshan D none of these ok first find number at first find number of ways of selecting a constant source of electric current desi Pashto total of five consonant 5C and choosing from them see this becomes and we noted and CR is equals to n factorial divided by and minus I effect tutorial in 25 Sector 10 / 3 factorial in 22 factory in which is equal to 10 ok number of ways for selecting and choosing vomit is equals to total of 44 CM choosing for selecting two from them to extend Sofia 482 in which is equal to after sex ok now total number of vowels and consonant total number of vowels and consonant poster tuval and 3 consonants this becomes 5 now number of ways the world there Piya request 25 factory because that have example as we have three letters select an and so I can arrange it as M L and also then an m the word that is formed from these two vowels and 3 consonants can also be arranged in five factories near perfect factories that now Dekho total number of edges request to this 56510 inter 2 means sex into this Five Factor is equals to wanted so total of 7200 MB can see that option this thank you No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! Solution The correct option is C72002 out of 4 vowels can be chosen in 4C2 ways and 3 out of 5 consonants can be chosen in 5C3 ways.Thus, there are (C2×5C3) groups, each containing 2 vowels and 3 consonants.Each group contains 5 letters that can be arranged in 5! ways.∴ Required number of words(4C2×5C3)×5!=60×120=7200How many words are formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants?=60×120=7200.
How many words of 3 consonants and 2 vowels can be formed out of 5 consonants and 3 vowels?Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters. = 5! = 120.
How many words of 3 consonants and 2 vowels can be formed?Number of groups, each having 3 consonants and 2 vowels = 210. Each group consist of 5 letters.
How many 5 letter words can be formed using 2 vowels and 3 consonants?There are 1440*10 = 14400 possible 5-letter “words” with two vowels and three consonants.
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