How many arrangements of the letters of the word DEFEATED ar

Question

How many arrangements of the letters of the word DEFEATED are there in which the three E are separated?

amit2k9 · Answer

different letters are

D - 2 times, E- 3 times, F,A and T - 1 times each = total 8

thus total permutation possible is = 8!/ 3!*2! = 56 * 60.

E - 3 times need to be separated

thus, total permutation - permutation with 3 E's together - permutation of 2 E's together = permutation with 3 E's separated.

permutation with 3 E's together = 6!( for 5 letters + 1 unit of 3 E's together) / 2! ( for 2 D's) = 6* 60
permutation for 2 E's together = 7!( for 6 letters + 1 unit of 2 E's together) / 2!

= 42 * 60.

hence required permutation = (56-48) * 60 = 480.

fluke · Answer

I am seeing this question as number of arrangements with no two E's together:

Total arrangements-Three E's together-Exactly Two E's together

Total arrangement=  \frac{8!}{3!2!}

Three E's together=  \frac{6!}{2!}

Exactly Two E's together= Two E's together - Three E's together =  \frac{7!}{2!}-\frac{6!}{2!}=6*\frac{6!}{2!}

\frac{8!}{3!2!}-\frac{6!}{2!}-6*\frac{6!}{2!}

\frac{8!}{3!2!}-\frac{7!}{2!}=\frac{7!}{2!}*\frac{1}{3}=840

Ans: "840"

Aj85 · Answer

hii fluke thanks for the solution. I did not know what to do after finding 3 e's together. But does this type of problem comes in GMAT ? haven't seen such a question in og and most people who score 700 + say they encountered only 1 combinatorics problem and that too based on general counting principle or probability. I am really confused whether to spend my time on important topics like inequality and number system or permutations and combinations.

fluke · Answer

I really don't think you will get this kind of problem on real GMAT. Inequality, number system, percent/ratio/mixture, statistics, geometry, other word problems; these should be the priority without a doubt. The following should be more than enough. It is advisable to go through these to increase your chances to get >50. math-combinatorics-87345.html math-probability-87244.html hardest-area-questions-probability-and-combinations-101361.html combined-probability-question-87294.html#p656120 ***************************************************************** If you can handle P&C from GPrep, you are good to go.

Aj85 · Answer

Thanks fluke once again for the reply and the links. I will go over the links from today itself.

zura · Answer

Hi Fluke, 
why can't we just subscrubt 7!/2! from 8!/3!2!?
i get the same answer
7!/2! - when 2 or 3 E are together

fluke · Answer

I did just that:

How did you arrive at this:
7!/2! - when 2 or 3 E are together

SVaidyaraman · Answer

Take the case with constraints placed at the leftmost

E-E-E---

The four blank places can be filled in 5!/2=60 ways.
 
1. The last E can be moved 3 more positions to the right in addition to its present position , each giving 60 ways for a total of 60*4=240 ways

2. Now push the next rightmost E also to the right. We have 60*3=180 ways. Pushing the two rightmost E's can continue till 60*2 and 60*1 ways i.e, a total of 360 ways

3. Pushing all the E's by 1 position, we have 60*3 and 60*2 and 60*1 i.,e 360 ways ways . Pushing all the three E's can continue till 60*2 and 60 *1 and 60*1 ways i.e, a total of 600 ways

The grand total is 240+360+600=1200

As a short cut 
step 1 is 60*4
Step 1 and 2 is 60*4 + (60*3 + 60*2 +60*1)
Step 1 ,2 and 3 is 60*4 + 60*3 + 60*2 +60 *1 + (60*3+60*2 +60*1) + (60*2+60*1) + (60*1)=1200

Thus these problems where we can place the constraints starting leftmost possible can be easily solved just by solving step 1.

outofpocket · Answer

Answer should be 1200.

D  E  F  E  A T  E  D

Q: ways E can be placed while separate from each other * ways rest of the alphabets can be distributed =

ways E can be placed while separate from each other:
_D_F_A_T_D_ ==> 6C3 = 20
ways rest of the alphabets can be distributed:
5 seats, 2 identical: 5!/2! = 60

A: 20 * 60 = 1200

any opinions?

KarishmaB · Answer

Whenever you need to separate out things, you should arrange the rest of the things first and then plug-in in the slots made. 
For example, you often have questions where you have say 6 boys and 6 girls and no two girls should together. First arrange the 6 boys in 6! ways. You have 7 slots then so you pick 6 and arrange the girls in those.

This question is very similar to that. 
DEFEATED has two Ds, an F, A and T. 
Arrange these 5 letters in 5!/2! = 60 ways such as DDFAT, FDDAT, DFATD etc

Now there are 6 spots to place the Es such that they are not together e.g. _ D _ D _ F _ A _ T _
Choose any 3 of these 6 in 6C3 = 20 ways and put the 3 Es there.

You get 60 * 20 = 1200 ways

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How many arrangements of the letters of the word DEFEATED ar

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How many arrangements of the letters of the word DEFEATED ar

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