Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 5)
5.
In how many ways can the letters of the word 'LEADER' be arranged?
Answer: Option
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
 Required number of ways = |
6! | = 360. |
| (1!)(2!)(1!)(1!)(1!) |
Video Explanation: https://youtu.be/2_2QukHfkYA
Discussion:
84 comments Page 2 of 9.
Dadasaheb Maske said:
7 years ago
Explain
(1)
Asam said:
7 years ago
Please explain this.
Manish said:
8 years ago
If vowels point is given then vowel to be treated as one-word & if that one word contains repeat letters they must be divided.
For eg- eeo.
3!/2!.
For eg- eeo.
3!/2!.
Manish said:
8 years ago
7 letters out of which 2 letters are repeated.
Therefore 7!/2!
Therefore 7!/2!
RDG said:
8 years ago
Sometimes vowel sometimes that way nothing could be understood please explain in sequence. And introduce the differences of combina and permu. Try to teach from a novice perspective.
Shwetha said:
8 years ago
Hello, @All.
In this question it means like how many ways (it means without repetition of letters) the "LEADER" letter can be arranged it supposed to be 6!=720. If the question is like how many different ways (with repetition of letters) can letter "LEADER" arranged then 6!/2! (2! is because of E letter repetition) = 360.
Hence the solution for this problem should be 720 as per my knowledge.
In this question it means like how many ways (it means without repetition of letters) the "LEADER" letter can be arranged it supposed to be 6!=720. If the question is like how many different ways (with repetition of letters) can letter "LEADER" arranged then 6!/2! (2! is because of E letter repetition) = 360.
Hence the solution for this problem should be 720 as per my knowledge.
Lavanya said:
8 years ago
Without mentioning why did they do with repetition.
Why can't we do without repetition?
Why can't we do without repetition?
Sanjib said:
8 years ago
LEADER : totally 6 letters. so= 6!.
And vowel 3 letters (2->E and 1->A) so we can solve in this way : 6! /2!=6*5*4*3*2! /2!=360 (ans).
And vowel 3 letters (2->E and 1->A) so we can solve in this way : 6! /2!=6*5*4*3*2! /2!=360 (ans).
Parthiban said:
8 years ago
LEADER : totally 6 words so= 6!.
L1, E2, A1, D1, R1 so= 1!*2!*1!*1!*1!.
6!=6*5*4*3*2*1=720.
1!*2!*1!*1!*1!= (1*1) (2*1) (1*1) (1*1) (1*1)=2.
Total words=720/2=360.
L1, E2, A1, D1, R1 so= 1!*2!*1!*1!*1!.
6!=6*5*4*3*2*1=720.
1!*2!*1!*1!*1!= (1*1) (2*1) (1*1) (1*1) (1*1)=2.
Total words=720/2=360.
(1)
Vamshi said:
8 years ago
They didn't ask without repetition? Then why you divided by 2?
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