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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  [#permalink]

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New post 18 Sep 2011, 14:05
3
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A
B
C
D
E

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0% (00:00) correct 100% (00:00) wrong based on 7 sessions

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How many arrangements of the letters of the word DEFEATED are there in which the three E are separated?
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Re: tough arrangement  [#permalink]

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New post Updated on: 19 Sep 2011, 01:44
1
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.
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Originally posted by amit2k9 on 19 Sep 2011, 01:13.
Last edited by amit2k9 on 19 Sep 2011, 01:44, edited 1 time in total.
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Re: tough arrangement  [#permalink]

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New post 19 Sep 2011, 01:19
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zura wrote:
How many arrangements of the letters of the word DEFEATED are there in which the three E are separated?


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"
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Re: tough arrangement  [#permalink]

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New post 19 Sep 2011, 01:32
fluke wrote:
zura wrote:
How many arrangements of the letters of the word DEFEATED are there in which the three E are separated?


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"



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.
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Re: tough arrangement  [#permalink]

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New post 19 Sep 2011, 01:57
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Aj85 wrote:
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.


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.
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Re: tough arrangement  [#permalink]

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New post 19 Sep 2011, 02:13
Thanks fluke once again for the reply and the links. I will go over the links from today itself.
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Re: tough arrangement  [#permalink]

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New post 26 Sep 2011, 05:49
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
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Re: tough arrangement  [#permalink]

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New post 26 Sep 2011, 08:47
zura wrote:
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


I did just that:

How did you arrive at this:
7!/2! - when 2 or 3 E are together
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How many arrangements of the letters of the word DEFEATED ar  [#permalink]

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New post 22 Nov 2016, 01:52
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zura wrote:
How many arrangements of the letters of the word DEFEATED are there in which the three E are separated?

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.
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Re: How many arrangements of the letters of the word DEFEATED ar  [#permalink]

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New post 27 Jul 2018, 10:03
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?
Re: How many arrangements of the letters of the word DEFEATED ar &nbs [#permalink] 27 Jul 2018, 10:03
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