Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 4)
4.
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Answer: Option
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
| = (7C3 x 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! |
| = 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
Discussion:
65 comments Page 1 of 7.
Anil said:
1 decade ago
Why multiply with 5?
Rahul said:
1 decade ago
@ 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.
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.
(1)
Student said:
1 decade ago
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 :(
7P3*4P2 = 2520
One zero less to match the answer of 25200 :(
Pranay said:
1 decade ago
The question doesn't mention "distinct" consonants or vowels. So repetition should be allowed.
Suchi said:
1 decade ago
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! ?
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:
1 decade ago
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:
1 decade ago
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.
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:
1 decade ago
Why we need multiply 120 with 210 is it necessary?
Navin said:
1 decade ago
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:
1 decade ago
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).
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).
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