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# out of 7 consonants and 4 vowels, how many words of 3 consonants and 3 vowels can be formed?16800

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## Out of 7 Consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Click here👆to get an answer to your question ✍️ Out of 7 Consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

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## Out of 7 Consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

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Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is A)

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) is given by

7 C 3 ​ × 4 C 2 ​ = (7−3)!3! 7! ​ × (4−2)!2! 4! ​ = 3×2×1 7×6×5 ​ × 2×1 4×3 ​ =210

Number 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.

205 11

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## Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 : Problem Solving (PS)

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? A. 210 B. 1050 C. 25200 ...

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## Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Answer (1 of 16): 7C3 X 4C2 X 5! = 35*6*120 = 25200

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

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Sort Nikhil Dewan

Studied Mathematics4y

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How many words with 3 consonants and 2 vowels can be made from 7 consonants and 4 vowels?

Clearly we need to select 3 consonants out of 7, hence it is clear to use C(7,3) and we need to have 2 vowels as well that clearly tells us about C(4,2).

Remember remember remember, just now we have only selections but different words means vowels and consonants can be arranged.

C(7,3) x C(4,2) x 5!

35 x 6 x 120 35200 Hope it helps. Related questions

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Out of 7 consonants and 4 vowels, how many words of 3 consonants, and 2 vowels can be formed?

7C3 * 4C2 = 35 * 6 = 210 groups of 5 such letters.

Then apply the Counting Principle to each group because order counts here.

5x4x3x2x1 = 120

So, 120 arrangements for each of the 210 groups = 120 * 210 = 25,200.

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Meha Bhalodiya

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Out of 8 consonants and 4 vowels, how many words can be made, each containing 4 consonants and 3 vowels?

No. of consonants available = 8

No. of vowels available = 4

No. of consonants to choose = 4

No. of vowels to choose = 3

No. of ways to choose consonants = 8C4 = 70

No. of ways to choose vowels = 4C3 = 4

No. of unique consonant-vowel sets = 70 x 4 = 280

Since the resulting word is 7 letters long, for each set of consonants and vowels, there can be 7! ways to arrange the letters to form a word.

So, the total number of words that can be formed = 280 x 7! = 280 x 5040 = 1,411,200

Hope it helps..!! Thank you❤️ Sanjiv Soni

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Out of 7 consonants and 4 vowels, how many words of 3 consonants, and 2 vowels can be formed?

Out of 7 consonants and 4vowels ,to form words of 3 consonants and 2 vowels.the number of ways

=7C3 × 4C2 ways

=[7!/3!(7–3)!] *[4!/2!(4–2)!]

=(7!/+3!*4!)*(4!/(2!*2!)

=35*6 =210

Number of ways of selecting 5 letters of which 3are consonants and 2 are vowels is 210

We can form a five letter word on 5! Ways=5*4*3*2*1=120ways

Total number of ways 120*210=25200 ways

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Out of 5 consonants and 3 vowels, how many words of 2 consonants and 2 vowels can be formed?

Ronak Jain

Studied at Malaviya National Institute of Technology, JaipurUpdated 7y

7C3 X 4C2 X 5! = 35*6*120 = 25200 Sponsored by Sirv

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Muntasir Jeio

Junior Executive (Event Management) at KUET Career Club (2018–present)4y

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How many words with 3 consonants and 2 vowels can be made from 7 consonants and 4 vowels?

choose any 3 consonants from the 7 by C(7,3) and 2 vowels from 4 by C(4,2). then Multiply them. now we have words having 3 consonants and 2 vowels but they can be arranged among them too. so again multiply it by 5! .so the ans is

C(7,3)*C(4,2)*5! Mathematics Answered by Enrico Gregorio

You can have the following ten configurations, listing first the cases where the two vowels are adjacent.

VVCCC CVVCC CCVVC CCCVV VCVCC VCCVC VCCCV CVCVC CVCCV CCVCV

Each configuration can be filled in

7 3 ⋅ 4 2 73⋅42

ways, if you allow repetitions of the letters, so 53880 total.

If you don’t allow repetitions of the letters, then the number for each configuration is

(7⋅6⋅5)⋅(4⋅3) (7⋅6⋅5)⋅(4⋅3) and so the total is 25200 25200