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 6 of 7.
Ankita said:
8 years ago
@Himatheja.
We have to make word from both consonants and vowel that's why we have to multiply combination of consonants with vowel also,
7c3*4c2.
= 35*6.
= 210.
We have to make word from both consonants and vowel that's why we have to multiply combination of consonants with vowel also,
7c3*4c2.
= 35*6.
= 210.
(1)
Aadesh said:
8 years ago
I agree @Ankita.
That as we are picking up the words by applying combination means that we are forming the words therefore answer. It Should be 210.
That as we are picking up the words by applying combination means that we are forming the words therefore answer. It Should be 210.
(1)
PROBODH said:
8 years ago
The correct answer should be option A.
(2)
Adithya UN said:
8 years ago
Here, the answer is 302400.
Because you should 3 constants from 7 and 2 vowels from 4 and multiply then the answer is (7*6*5) * (4*3).
Later you should arrange them by multiplying with 5!.
Because you should 3 constants from 7 and 2 vowels from 4 and multiply then the answer is (7*6*5) * (4*3).
Later you should arrange them by multiplying with 5!.
Burt Yellin said:
8 years ago
I'm not sure this is the correct forum for this question, but here goes. If I have 4 consonants and there are 5 vowel sounds that can be pronounced hard or soft, how many different pronunciations are possible?
If this is the wrong forum can you direct me to a place that could answer that question?
If this is the wrong forum can you direct me to a place that could answer that question?
Gezoi said:
8 years ago
7P3 * 5P2 =2520.
Then considering combining them accounting for repetitions we get = 5!/(3!*2!)=10.
Therefore =2520*10=25200.
Then considering combining them accounting for repetitions we get = 5!/(3!*2!)=10.
Therefore =2520*10=25200.
(1)
Mark Burnham said:
8 years ago
This is a question of permutation, as the order in which letters are arranged changes the final result.
Therefore = 7P3*4P2 = 2520.
However, there are 10 different ways that the consonants and vowels can be arranged.
So, the answer is 2520*10,
=2520.
Therefore = 7P3*4P2 = 2520.
However, there are 10 different ways that the consonants and vowels can be arranged.
So, the answer is 2520*10,
=2520.
May said:
8 years ago
Why do 5!?
What if the consonants or vowels have repetition? Does it necessary to multiply with 5!?
What if the consonants or vowels have repetition? Does it necessary to multiply with 5!?
(5)
K ANIRUDH said:
8 years ago
Hi, @All.
I am just confused with committee problems and this question in committee problems we are not doing this 5! thing if a committee of 5 people is to be made, we are just selecting the men and women out of given men and women and multiplying why it is not similar in this case.
Can anyone help to understand the problem? Please help me.
I am just confused with committee problems and this question in committee problems we are not doing this 5! thing if a committee of 5 people is to be made, we are just selecting the men and women out of given men and women and multiplying why it is not similar in this case.
Can anyone help to understand the problem? Please help me.
(5)
Rahul said:
7 years ago
Why we want to multiply with 5! Please tell me.
(8)
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