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from 6 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels?

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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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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.

Hence, the answer is 25200.

205 11

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Permutation and Combination

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Discussion :: Permutation and Combination - General Questions (Q.No.4)

Permutation and Combination - Important Formulas

«« Permutation and Combination - General Questions

4.

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 [D]. 21400 [E]. None of these Answer: Option C Explanation:

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

= (7C3 x 4C2)

= 7 x 6 x 5 x 4 x 3

3 x 2 x 1 2 x 1 = 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 x 4 x 3 x 2 x 1 = 120.

Required number of ways = (210 x 120) = 25200.

Video Explanation: https://youtu.be/dm-8T8Si5lg

Workspace Report

Anil said: (Oct 4, 2010)

Why multiply with 5?

Rahul said: (Oct 28, 2010)

@ Anil: The basic idea here is that first you pick out 3 consonants out of 7 and 2 vowels out of 4. Then you take the 5 letters that you have and make a word out of it. To pick out 3 consonants out of 7 you use the combination without repetition formula which states n!/(n-r)!r! which gives you 7C3 and the same for vowels gives you 4C2.

Once you have done that, you still need to make a word of it. You have 5 letters now that can be arranged into 5! ways to make a word so you multiply the three values since each step is dependent on the step before to make the final word. The answer comes out to be 35 x 6 x 120 respectively for each step mentioned chronologically.

Student said: (Mar 18, 2011)

Can't we apply permutation directly here since in this question arrangement of words is to be done. i.e.

7P3*4P2 = 2520

One zero less to match the answer of 25200 :(

Pranay said: (Jul 18, 2011)

The question doesn't mention "distinct" consonants or vowels. So repetition should be allowed.

Suchi said: (Aug 9, 2011)

c = consonants v = vowels

Why has the num of letters has to be 5?

Can't it be 4c+3v(7 letter) or 5c+4v(9 letters)

Then how can v take 5! ?

Machender said: (Aug 22, 2011)

Miss suchi, from given question it was clear that out of 7 consonants 3 consonants and 2 vowels of 4 vowels could considered.

M.V.KRISHNA/PALVONCHA said: (Sep 10, 2011)

Hi suchi. Given 3c out of 7c 2v out of 4v.

So, required number of letters are 5. they can be arranged among

themselves in 5!.

Hope you understood.

Student said: (Nov 1, 2011)

Why we need multiply 120 with 210 is it necessary?

Navin said: (Dec 18, 2011)

Hey they anly asked us how many ways to form the letters they didint ask how many ways to arrange it the answer is wrong it should be 210.

Sachin Jain said: (Jan 7, 2012)

Dear friends

Keep in mind first that it is a permutation problem becoz the order of letters matter in order to make different words. Secondly out of 7 consonants we have to select 3 so when we select a consonant out of 7 it will not again be considered to be selected so the repetition is not allowed. Now the formula for Permutation without repetition is (n!)/(n-r)!. So it will be 7P3*4P2.

=(7!/4!)*(4!/2!)=2520 (not 25200).

Litu said: (Mar 9, 2012)

Hi,

I do agree with Sachin Jain as it is a permutation query. We can arrange it in any way using using 3 consonants out of 7 & 2 vowels out of 4.

Ex: If there is bcdfghj then we can do it bcd n also cbd so it is permutation type not combination so the answer should be 7P3 * 4P2 = 2520

But again we have to arrange these 2 groups together. So I'm not getting form this point.

Can anyone clear the same?

Rizwana said: (Jun 21, 2012)

Hi,

I always find problem in using C and P in formula I could not distinguish when to use nCr and when to use nPr please help in finding difference.

Benni said: (Jun 25, 2012)

How can we find permutation and combination problem ?

Veyron999 said: (Aug 31, 2012)

This question can be solved by two methods

1) using combinations:

we have to select 3 consonants from 7 and 2 vowels from 4,

we will use 7C3 x 4C2, this implies that for eg. consonants are B C D F G H J, then

if BCD is there then CBD and DBC won't be there and so on..

but they will make different words, so

we arrange them later on by multiplying by 5! ..

eg. one combination will be BCD AE this can be arranged in 5! ways and so on...

so ans. is 7C3 x 4C2 x 5! = 25200

2) using permutations:

unlike previous method if we use 7P3 we also consider BCD, CBD and all possible arrangements...

then we get combinations of the type CCCVV (C- consonant, v- vowel)

but they can be arranged in 5! ways eg- CVCVC, VVCCC, etc..

but using permutation, we have already accounted for those combinations in which all 3 consonants occur together and 2 vowels occur together.. so to eliminate those cases, we need to divide it by the repetitions i.e. divide by 3!2! (for consonants and 2 for vowels) and then multiply this with the initial result..

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Out of 7 consonants and 4 vowels, how many words o

Permutations and Combinations Questions & Answers : Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

308 Q:

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

A) 25200 B) 52000 C) 120 D) 24400 Answer:   A) 25200 Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (

7 C 3 7C3 * 4 C 2 4C2 ) = 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! = 120

Required number of ways = (210 x 120) = 25200.

Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability

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