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:
(Wed, Sep 8, 2010 10:45:45 AM)
Hmm Krishna you are right. I got the answer by your way.
Sunil said:
(Tue, Nov 16, 2010 10:38:02 AM)
Explain Please
Sam said:
(Sun, Feb 13, 2011 05:11:03 AM)
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:
(Sat, Apr 9, 2011 05:18:02 AM)
6*5*4*3*2*1=720
720/2=360
Umakant said:
(Thu, May 12, 2011 05:45:15 AM)
Why is wrong 5x4x3x2x1?
Umakant said:
(Thu, May 12, 2011 05:46:30 AM)
In this Method
6*5*4*3*2*1=720
720/2=360
Why divide by 2
Please explain it
Praveen said:
(Tue, May 31, 2011 10:38:02 PM)
We can't divide by 2 may be what not 5!?
Sonu said:
(Thu, Jul 14, 2011 02:22:51 AM)
Ritu explain it in a very gooo way thanks a lot ritu.
Vinny said:
(Mon, Jul 18, 2011 11:12:17 AM)
LEADER contains 6 letters.
So 2 are similar letters.
Therefore 6!/2!=360.
Poornima.R said:
(Fri, Aug 26, 2011 06:32:17 PM)
Why we want to divide this 720 by 2?
Achutharaj said:
(Wed, Sep 28, 2011 10:14:33 AM)
Hi poornima,
E is repeated in two times so we divided
leader -l,e,a,d,r ( e is repeted 2 times so considered single time)
6!=6x5x4x3x2x1/2x1=360
Tezan said:
(Fri, Sep 30, 2011 11:34:56 PM)
Why we take 6!/2! Whay we have not take vowel 3 and consonent 3. like (3+1)4
4!= 4*3*2*1= 24
3!/2!=3
24*3= 72
Lalit said:
(Mon, Jan 23, 2012 11:49:25 PM)
Helo frndz in this word LEADER thre r 6 leters in which 2 leters are repeted thats why we divide it by 2! and the ans is 6!/2!
Tanaya said:
(Wed, Feb 1, 2012 04:42:46 PM)
I have query that all these problems belong to combination and not permutation section?
Abhishek said:
(Mon, Sep 17, 2012 09:34:47 PM)
Why not 5! * 2!? because E is repeated so now consider 5 letters so 5! and E is repeated twice so 2! ??
Sujit said:
(Tue, Nov 27, 2012 05:39:27 PM)
@Tanaya: both permutation and combination are used in this section.
In some problems we arrange the letters means they are of type permutation..and in some cases like selection group we used combination :-).
@Abhishek: let me explain you with the example
Suppose your name is : AAB which contain A 2 times (in above question E repeate 2 times so i consider name with one character repeat twice)
AAB can be arranged in
AAB ( 1)
ABA (2)
BAA (3)
Means 3 ways..
This ans can be calculate directly using formula
Formula:
(number of characters)!
------------------------
(number of repeat characters) !
In AAB there are 3 characters and one character repeat twice
So above formula become
3! 3*2*1
-- = ------ = 3 which we got above by doing manually
2! 2*1
similarly in above question
LEADER contains 6 character and E repeat twice so using formula
6!
--- = 360.
2!
Hope you got this :-).
Jonathan said:
(Thu, Dec 6, 2012 01:53:01 PM)
LEADER
One way of arranging this is
"EADERL" (The first 'E' and the second 'E')
If I want to put the second 'E' in place of the first 'E' to make another arrangement, I would get
"EADERL" which is the same as the previous one.
POINT: we have shown that eventhough I interchange the first 'E' and the second 'E', we came up with the same arrangement.
MEANING... if there are TWO same letters, there will be DUPLICATION...
That is the reason why we DIVIDE by TWO!
It's not 5! because still, there are six letters we are arranging!
Hope this insight can help you!:)
Sundar said:
(Fri, May 3, 2013 01:19:44 PM)
In how many different ways can the letters of the word 'ORANGE' be arranged so that the three vowels never come together ?
Answer :
The total number of ways of arranging ORANGE = 6!
The total number of ways o,a and e can be arranged = 3!
The total number of groups when the vowels are one group and the rest are individuals= 4!
= 6! - 4! x 3!
= 576
In how many different ways can the letters of the word 'EXTRA' be arranged so that the two vowels never come together ?
Answer:
The total number of ways of arranging EXTRA = 5!
The total number of ways e and a can be arranged = 2!
The total number of groups when the vowels are one group and the rest are individuals XTR(EA) = 4!
= 5! - 4! x 2!
= 72
Required number of ways =