Aptitude - Permutation and Combination - Discussion

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"Life is like riding a bicycle. To keep your balance you must keep moving."

- Albert Einstein

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

Answer: Option A

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.

Rahul said: (Thu, Oct 28, 2010 12:46:55 AM)

@ 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: (Fri, Mar 18, 2011 10:40:24 PM)

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: (Mon, Jul 18, 2011 08:23:01 PM)

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

Suchi said: (Tue, Aug 9, 2011 09:38:33 PM)

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: (Mon, Aug 22, 2011 11:12:20 PM)

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: (Sat, Sep 10, 2011 10:08:54 AM)

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: (Tue, Nov 1, 2011 10:00:21 AM)

Why we need multiply 120 with 210 is it necessary?

Navin said: (Sun, Dec 18, 2011 06:16:13 PM)

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: (Sat, Jan 7, 2012 04:27:38 PM)

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 7P34P2. =(7!/4!)(4!/2!)=2520 (not 25200).