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Math Expert V
Joined: 02 Sep 2009
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If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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10 00:00

Difficulty:   45% (medium)

Question Stats: 69% (01:35) correct 31% (01:41) wrong based on 375 sessions

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If the 6 letters in the word NEEDED are randomly rearranged, what is the probability that the resulting string of letters will not be in alphabetical order?

A. $$\frac{1}{720}$$

B. $$\frac{1}{120}$$

C. $$\frac{1}{60}$$

D. $$\frac{59}{60}$$

E. $$\frac{719}{720}$$ This question was provided by Crack Verbal for the Game of Timers Competition _________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59236
Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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Bunuel wrote:
If the 6 letters in the word NEEDED are randomly rearranged, what is the probability that the resulting string of letters will not be in alphabetical order?

A. $$\frac{1}{720}$$

B. $$\frac{1}{120}$$

C. $$\frac{1}{60}$$

D. $$\frac{59}{60}$$

E. $$\frac{719}{720}$$ This question was provided by Crack Verbal for the Game of Timers Competition OFFICIAL EXPLANATION: FROM CRACK VERBAL:

The word NEEDED arranged Alphabetically is DDEEEN.
We can use 1 – (Probability of not getting an alphabetical order) that will give Probability of getting an alphabetical order.
So, P(getting an alphabetical order) = 1 – (P(not getting an alphabetical order))
= 1 – (Not getting DDEEEN).
= 1 – ((2/6)(1/5)(3/4))(2/3)(1/2)(1/1))
= 1 – 1/60 = 59/60.
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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5
1
The alphabetical order is DEN.
Now total number of ways to arrange the letters is 6!/(2!*3!) = 60
Only in one way the letters will be in alphabetical order. DDEEEN.
##### General Discussion
Manager  G
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If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
We can get NEEDED in alphabetic order only in one case DDEEEN (note that since DD and EEE are identical, only one case is possible, not like D1D2E1E2E3N OR D2D1..... and so one). Now, let's calculate denominator $$\frac{6!}{2!(two DDs)3!(three Es)}$$ = 60. Since only in case we can get alphabetical order out of 60, in 59 cases out of 60, we get not alphabetical order, thus $$\frac{59}{60}$$. Our answer is D

Originally posted by mira93 on 05 Jul 2019, 08:08.
Last edited by mira93 on 06 Jul 2019, 02:16, edited 3 times in total.
Senior Manager  P
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
1
If the 6 letters in the word NEEDED are randomly rearranged, what is the probability that the resulting string of letters will not be in alphabetical order?

A. 1/720

B. 1/120

C. 1/60

D. 59/60

E. 719/720

only one way we can arrange the word in alphabetical order: DDEEEN
total number of cases: 6!/(2!*3!) = 60

so the probability of alphabetical order = 1/60
so the probability of not alphabetical order = 1- (1/60) = 59/60
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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it will be 1/60

One way to arrange the letters in alphabetical order and 6!/3!x2! ways in total.
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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There are 6!/2!3!=60 ways of arranging NEEDED

If the letters are in alphabetic order, it must be DDEEEN which is the only permutation

So the probability of getting an arrangement in alphabetic order is 1/60

This means the probability that resulting string of letters will not be in alphabetical order = 1 - 1/60 = 59/60

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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
If the 6 letters in the word NEEDED are randomly rearranged, what is the probability that the resulting string of letters will not be in alphabetical order?

N-1, E-3 & D-2
Now arrangements = 6!/3!/2! = 60
EEEDDN is the string in alphabetical order
#of favourable outcomes = 60 - 1 = 59 since all other strings will NOT be in alphabetical order
#of total outcomes = 60

Probability that the resulting string of letters will not be in alphabetical order = 59/60

IMO D
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
Total ways of forming the words
= 6!/3!2!
= 60

the only words are in alphabetical order is
DDEEEN

that is only one way of forming the words.. 1
hence the number of ways of not forming the letter in alphabetical order is 60-1 = 59
Hence total probability is 59/60
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If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
There is one way of choosing th letters in alphabetical order which is DDEEEN.

Total no of ways = 6!/2!3! = 60
Total 60 ways

So the answer should be 1/60

Option C.

Posted from my mobile device
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Originally posted by prashanths on 05 Jul 2019, 08:19.
Last edited by prashanths on 06 Jul 2019, 01:14, edited 1 time in total.
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
Total arrangements possible : $$\frac{6!}{3!*2!}$$ = 60

(3! because letter E is repeated 3 times & 2! because letter D is repeated 2 times while total letters are 6 so 6!)

Possible arrangements in alphbetical order : 1 (DDEEEN)

Probability of arrangement of letters in alphabetical order: $$\frac{favorable outcomes}{total possible outcomes}$$ = $$\frac{1}{60}$$

So, inverse probability (which is what we are asked to find) = 1 - $$\frac{1}{60}$$ = $$\frac{59}{60}$$

Hence, Ans should be (D)
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
Total no of ways of arranging the letters
6!
—. =60
3!*2!

Ways to arrange Letters in alphabetical order=1

Probability that result is not in alphabetical order is 1- 1/60= 59/60

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Manager  G
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
we have to arrange needed
so standard formula
N!/p!*Q! where p and q are the number of identical objects in set N

so NEEDED : 6 letters , E- 3 times D-2 times

so total arrangements without any condition = 6!/ (2!*3!)= 60

required probability : favorable cases /total cases = favorable /60

so we can eliminate A,B,and E

now if confused just look at options C and D the are complimentary to each other P(R)=1- p(R not)

now arranging the string alphabetically has only 1 way for sure so its complement will be the required probability

P(Required)=1- probability of arranging alphabetically = 1-1/60= 59/60
therefore D
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
D.

Total no of words that can be formed is 6!/(2!*3!) = 60
There's only one way to write the letters in alphabetical form, so 59 ways to not do so.

Probability of not doing so: 59/60
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GMAT 1: 640 Q45 V35 GMAT 2: 660 Q48 V33 Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
Two Methods

1) Needed - can be arranged in exactly one way in alphabetical manner - DDEEEN when we consider D and E identical . Total possibilities considering them similar = 6!/(2!*3!) = 60 ways. Probability 1/60.

2) Needed - can be arranged in alphabetical manner in DDEEEN where there is 2*1*3*2*1 ways for arrangement in the same manner (consider D and E unique) = 12 and Total possibilities considering them unique will result in 6! ways = 720. Probability = 1/60.

For this question both Unique and non-unique works - IMO C
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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2
Hi,
The word is NEEDED.
We can rearrange the word in the following way:
$$\frac{(6!)}{(3! * 2!)}$$= 60

Now if it follows the alphabetical order then the word is: DDEEEN (Only one possible value)
So, the probability for not in alphabetical order is :

1 - $$\frac{(1)}{(60)}$$
=> $$\frac{(59)}{(60)}$$

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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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1
Total number of ways of arranging NEEDED = 6!/(2!*3!) = 720/(2*6) = 720/12 = 60

Number of ways in which they are in ascending order = 1 (DDEEEN)
--> Number of ways in which they are NOT in ascending order = 60 - 1 = 59

--> Required Probability = 59/60

IMO Option D

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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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2
6 things can be arranged in a straight line in 6! ways. Among 6 things E is repeated 3 times and D is 2 times so the thing of actual arrangement is 6!/(2!*3!) = 60.
Now we need the item not to be the way as it is written so it means we have to know the ways in which Needed can be written. i.e. 1 way.
so probability is 59/60.
ans=D
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Re: If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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2
NEEDED

Alphabetical order = DDEEEN

We will calculate prob of getting alphabetical order and then subtract it from 1 to get prob of not getting alphabetical order

Denominator =Total number of ways (permutations) 6 letters can be arranged
=6*5*4*3*2*1

Numerator = permutations with restrictions for alphabetical order... 1st 2 must be D, next 3 must be E

=(2*1)*(3*2*1)*(1)

prob = Numerator/Denominator = 2*3*2/6*5*4*3*2

This can be simplified to 1/60 which is prob of getting alphabetical order

prob of NOT getting alphabetical order is 1-1/60 = 59/60

Ans: D
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If the 6 letters in the word NEEDED are randomly rearranged, what is  [#permalink]

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2
needed can be arranged in 6! / 3! * 2! which is 60 ways

out of which alphabetical order arrangement is 1,
so not in alphabetical is 59/60

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thanks If the 6 letters in the word NEEDED are randomly rearranged, what is   [#permalink] 05 Jul 2019, 08:36

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