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 4 of 9.
Umakant said:
1 decade ago
In this Method
6*5*4*3*2*1=720
720/2=360
Why divide by 2
Please explain it
6*5*4*3*2*1=720
720/2=360
Why divide by 2
Please explain it
Madhu said:
2 decades ago
Explain this please?
Abhishek krishna said:
1 decade ago
Hi Madhu and Shweta, the answer is right if see I tell how the letter E is repeated two times so it has to be divided and rest of letters are only one times occurred, so dividing from 1 did not make any thing.
Neeraj said:
1 decade ago
Hmm Krishna you are right. I got the answer by your way.
Sunil said:
1 decade ago
Explain Please
Sam said:
1 decade ago
Leader has 6 letters so it can be arranged in 6! ways but letter e is repeated twice so ans is 6!/2!=360.
Ritu said:
1 decade ago
6*5*4*3*2*1=720
720/2=360
720/2=360
Umakant said:
1 decade ago
Why is wrong 5x4x3x2x1?
Yuvraj k said:
7 years ago
I think the Answer is 720.
Sandeep said:
9 years ago
I have a query.
LEADER= (LADR) +) (EE) = 4 + 1 = 5! = 120, and divided it by 2C2!, why can't be the answer 120.
The same way for CORPORATION problem is solved. All vowels are a set & consonants are of another set and as the vowels are scrambled it was divided by vowels combinations.
LEADER= (LADR) +) (EE) = 4 + 1 = 5! = 120, and divided it by 2C2!, why can't be the answer 120.
The same way for CORPORATION problem is solved. All vowels are a set & consonants are of another set and as the vowels are scrambled it was divided by vowels combinations.
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