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 3 of 9.
Isha said:
9 years ago
It is 720.
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).
Shivaraj kavalaga said:
9 years ago
@Vikram.
EYE
It contains 3 letters i.e 3!
Now take reapeted letters i.e E(2times repeated)=2!,
EYE = 3!/2!,
= 3 * 2 * 1/2 * 1,
= 3.
EYE
It contains 3 letters i.e 3!
Now take reapeted letters i.e E(2times repeated)=2!,
EYE = 3!/2!,
= 3 * 2 * 1/2 * 1,
= 3.
Shivaraj said:
9 years ago
LEADER
It contains 6 letters i.e=6!
Now take repeated letters i.e E(2 times), so it can be divide by 6!
i.e LEADER =6!/2!.
= 6 * 5 * 4 * 3 * 2 * 1/2 * 1
= 360
So,360 is correct answer.
It contains 6 letters i.e=6!
Now take repeated letters i.e E(2 times), so it can be divide by 6!
i.e LEADER =6!/2!.
= 6 * 5 * 4 * 3 * 2 * 1/2 * 1
= 360
So,360 is correct answer.
Vikram said:
9 years ago
Take the example of the word EYE:
It can be arranged in 3 ways only , namely EYE , EEY, YEE as the letter E is repeated twice.
So the answer is not 3! but 3!/2!.
Applying the same for LEADER, we have the answer 6!/2!.
It can be arranged in 3 ways only , namely EYE , EEY, YEE as the letter E is repeated twice.
So the answer is not 3! but 3!/2!.
Applying the same for LEADER, we have the answer 6!/2!.
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.
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