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 7 of 9.
Prakash said:
9 years ago
Why we use permutation formula in this example?
Is there any way to solve this?
Is there any way to solve this?
Gowtham said:
8 years ago
@Khan.
The word SAMPLE has 6 letters so we apply the formula in 6!= 720 ways can we arranged.
Am I Right?
The word SAMPLE has 6 letters so we apply the formula in 6!= 720 ways can we arranged.
Am I Right?
Shekhar said:
8 years ago
In how many ways can the letters of the word 'DIRECTOR', be arranged so that the vowels are never together? Please solve this.
Ezaz Ahmed said:
8 years ago
Thanks for giving the explanation of the answer.
Vamshi said:
8 years ago
They didn't ask without repetition? Then why you divided by 2?
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)
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).
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?
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.
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.
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