If the 6 letters in the word NEEDED are randomly rearranged, what is

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

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

Answer is (D)

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

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 - \(\frac{(1)}{(60)}\)
=> \(\frac{(59)}{(60)}\)
IMO Answer is: D

Please hit kudos if you like the solution.

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

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

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

which is the answer D

total combination "needed" 6!/(3!2!) = 60

"den" is the one and only alphabetical order = 1

therefore,
probability of not in order = total probability- probability in order
= 60/60 - 1/60
=59/60

Posted from my mobile device

Total number of arrangements of the alphabets of 'NEEDED'= 6!/3!*2!= 60
Number of ways in which the resulting string of letters will not be in alphabetical order= 60-1=59

Probability = 59/60

IMO D

We need to find the probability of the resulting string of letters (NEEDED) not to be in alphabetical order

So we can find this by : P(all) - P(alphabetical order) = 1- P(DDEEEN)

Total no of ways of arranging the letters = \(\frac{6!}{2! * 3!}\) = 60
No of ways of arranging the letters in alphabetical order = 1 ( there is only one way to arrange the letters in alphabetical order)

P(Alphabetical order) = \(\frac{1}{60}\)

P(Not Alphabetical order) = 1- \(\frac{1}{60}\) = \(\frac{59}{60}\)

The Answer is D.

Consider +Kudos if this helps.

Question: 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?

Permutations with duplicates formula will be needed to determine how many letter combinations are possible.
\(n\) = NEEDED = 6; The word "needed" has a total of 6 objects.
\(n_1\) = E = 3; E is repeated 3 times.
\(n_2\) = D = 2; D is repeated 2 times.

\(\frac{(n!)}{(n_1!)(n_2!)} = \frac{(6!)}{(3!)(2!)} = \frac{(6 \times 5\times4\times3\times2\times1)}{(3\times2\times1)(2\times1)} = \frac{(6\times5\times4)}{(2\times1)} = \frac{(6\times5\times2)}{(1\times1)} =\) [60] \(\implies\) Total combinations possible without regard to alphabetical order.

Next, we will work the probability. The easiest way to process this would be to work with the opposite of what is asked in the question. The question asks for the probability of the letters NOT being in alphabetical order. However, it would be easier to subtract the number of combinations that ARE in alphabetical order from the total.

Combinations in alphabetical order = DDEEEN = [1]; (basically, there are no other ways to put these letters in alphabetical order... just think about it for a second.)

[Total combinations possible without regard to alphabetical order] \(-\) [Combinations in alphabetical order]
\(=\) [Combinations NOT in alphabetical order]
[60] \(-\) [1] = [59]

Correct Answer: D \(\implies\) \(\frac{59}{60}\) \(\implies\) \(\frac{\text{ [Combinations NOT in alphabetical order]}}{\text{ [Total combinations possible without regard to alphabetical order}]}\)

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