tough arrangement (experts pls help)

To Solve the arrangement,first we should get the total number of arrangements possible in DEFEATED

A)Total Arrangements : 8!/3!*2!
3! :refers to the occurences of E 3 times
2!:refers to the occurences of D 2 times

B)Getting occurences when E is together: 7!/2!
for any pattern like EE(------), we will have 7! ways.dividing by 2 as D comes 2 times.This will cover both the scenarios when 2 or more E come together.

Subtracting B from A,we will get 840.

mail.SuhailSharma wrote: B)Getting occurences when E is together: 7!/2!
for any pattern like EE(------), we will have 7! ways.dividing by 2 as D comes 2 times.This will cover both the scenarios when 2 or more E come together.

Subtracting B from A,we will get 840.

Suhail, could you please explain how you got 7!/2!.
If you are taking Es together then you should have 6!/2!.

If you have taken just 2 'E's together then 7! makes sense. But then we should have 7!/(2!*2!). This is to take care of the combination which 2 'E's make with the single 'E'.

I went by another approach, but got different answer...Could anyone pls explain, where I went wrong?

DEFEATED after removing the 'E's can be written as

| D | F | A | T | D |

The '|' represents the spaces where an 'E' can come. There are totally 6 spaces were 3 'E's can come - in 6C3 = 20 ways.

Now the remaining letters can be arranged in (5!)/(2!) = 60 ways - (since we have 5 letters out of which 2 are same.)

So total number of options will be 60 * 20 = 1200 ways.

To Solve the arrangement,first we should get the total number of arrangements possible in DEFEATED

A)Total Arrangements : 8!/3!*2!
3! :refers to the occurences of E 3 times
2!:refers to the occurences of D 2 times

B)Getting occurences when E is together: 7!/2!
for any pattern like EE(------), we will have 7! ways.dividing by 2 as D comes 2 times.This will cover both the scenarios when 2 or more E come together.

Subtracting B from A,we will get 840.

If all the E's are taken together as one, then we have only 6 alphabets ( 2Ds,F,A,T and one grp of Es).

The no. of possible arrangements would be 6!/2!
so no. of arrangements in which 3Es aren't together = (8!/3!*2!) - (6!/2!)
= 3000

Where am I going wrong???

bmlaud wrote:

If all the E's are taken together as one, then we have only 6 alphabets ( 2Ds,F,A,T and one grp of Es).

The no. of possible arrangements would be 6!/2!
so no. of arrangements in which 3Es aren't together = (8!/3!*2!) - (6!/2!)
= 3000

Where am I going wrong???

There are cases where 2 'E's are together and one E is separate. We have to substract that conditions too. Which is given by 7!/(2!*2!). But another catch is that this will take care of the conditions when EEE comes together. So 6!/2! is not required!
But this wont yield you 840!! So i am stuck as well. Could you please check on my approach and help me with it?

To Solve the arrangement,first we should get the total number of arrangements possible in DEFEATED

A)Total Arrangements : 8!/3!*2!
3! :refers to the occurences of E 3 times
2!:refers to the occurences of D 2 times

B)Getting occurences when E is together: 7!/2!
for any pattern like EE(------), we will have 7! ways.dividing by 2 as D comes 2 times.This will cover both the scenarios when 2 or more E come together.

Subtracting B from A,we will get 840

For B:I got 7!/2!.............explanation below

I have to find the number of times we can have combination when 3E's are not together. So, i got the number when one or more E are together

example: (EE) {EDFATD}
The 6 alphabets in the curly braces, I consider (EE) as 1.So, the number comes to 7(Which can hhave 7! combinations).
Also out of the 7,D comes twice.....so total number of times a number formed witj E along with one or more will be
==7!/2!

Subtracting(A-B)
(8!/3!*2!-7!/2!)==( 3360-2520)==840

I hope the explanation is sufficient....

mail.SuhailSharma wrote: example: (EE) {EDFATD}
The 6 alphabets in the curly braces, I consider (EE) as 1.So, the number comes to 7(Which can hhave 7! combinations).
Also out of the 7,D comes twice.....so total number of times a number formed witj E along with one or more will be
==7!/2!

Suppose we substitute EE with X, then we have (X) {EDFATD} and as u said 7!/2! will give you the arrangements. But in this arrangement XEDFATD and EXDFATD are different!!
But in Actual case EEEDFATD and EEEDFATD are same. So you have to take in to consideration that as well!

mariah, could you please let us know the source?

If the question is about arrangement such that no two E's are together, then without any doubt the answer must be 1200.

Given word DEFEATED.

We have to arrange such that no two E's are together i.e all the three E's are separated.

First keep these 3 E's aside and arrange the remaining D, F, A, T, D. This can be done in 5!/2 = 60 ways.

Now once we arrange these 5 letters there will be 6 places (4 in between the letters and 2 at the extreme ends). Now we can place the 3 E's in any of these 6 palces in 6C3 ways = 20 ways.

So the total number of ways is 1200.

Folks, this question is similar to something like arranging 5 boys and 3 girls in a row such that no two girls are together.....

i have collected it from net. and oa may be inaccurate!

Suppose we substitute EE with X, then we have (X) {EDFATD} and as u said 7!/2! will give you the arrangements. But in this arrangement XEDFATD and EXDFATD are different!!
But in Actual case EEEDFATD and EEEDFATD are same. So you have to take in to consideration that as well!

Thanks Alara,

woke up from my slumber
As us said, we substitute EE with X. After that as XE and EX are same.
We will have the combination as 7!/2!*2!==1260.
which on subtracted from 8!/3!*2! will give==2100

let me know if further queries come up