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.
Anil said:
9 years ago
Total letters-6!
And repeated letters 2 so-2!
Then 6!/2! = 360 right answer.
And repeated letters 2 so-2!
Then 6!/2! = 360 right answer.
Nanda said:
9 years ago
Why shouldn't the vowels and consonants be grouped differently ? like LDR (EEA) which makes it 3 + 1 = 4!
And 3!/2! since E has repeated twice like in the previous problems.
And 3!/2! since E has repeated twice like in the previous problems.
Kumar said:
9 years ago
LEADER has 6 words totally.
But letter E repeatedly two times then, to get a required solution we use 6! (total words) /2! (repeated word). So, the answer is 360.
But letter E repeatedly two times then, to get a required solution we use 6! (total words) /2! (repeated word). So, the answer is 360.
SANJAY P said:
9 years ago
I have to plan audit in such a way that 24 activities are to be assessed in 6 different laboratories in a year with 9 assessors. Two assessors are required for on activity.
Manoj D Desai said:
9 years ago
Please See the question once carefully, it's just asked for number of ways so word repetition is not considered so the correct answer is 720 (6!) ,
If the question is asked like " How many DIFFERENT ways word Leader can be arranged" then it would have been 6!/2 for E-letter repeating 2 times.
If the question is asked like " How many DIFFERENT ways word Leader can be arranged" then it would have been 6!/2 for E-letter repeating 2 times.
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.
Dipankar Rabha said:
9 years ago
How did 360 come?
What is (1!)? What is the meant of this symbol? And can anyone provide the easy method for this type of problem?
What is (1!)? What is the meant of this symbol? And can anyone provide the easy method for this type of problem?
Divya said:
9 years ago
@Savitri.
It is;
L = 1.
E = 2.
A = 1.
D = 1.
R = 1.
It is;
L = 1.
E = 2.
A = 1.
D = 1.
R = 1.
Savitri said:
9 years ago
(1!)(2!)(1!)(1!)(1!)
How it is? Explain me.
How it is? Explain me.
Savitri said:
9 years ago
@Sundar, Can you please explain this, how it is 72?
= 5! - 4! x 2!.
= 72.
= 5! - 4! x 2!.
= 72.
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