In how many ways can letters the word ATTITUDE be rearranged such that

Question

In how many ways can letters the word ATTITUDE be rearranged such that no two Ts are adjacent to each other?

A. 1800
B. 2400
C. 3600
D. 4320
E. 6720

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tam87 · Accepted Answer

without the Ts, there are 5 other distinct letters: A I U D E
these 5 letters can arrange themselves in 5! or 120 ways

now, to ensure that no two Ts are together, there must be at least one letter between any two Ts . If we represent each of the other letters as X, then the arrangement can be 
_X_X_X_X_X_
here the three Ts can take any 3 of the 6 vacant positions, and this will also ensure no two Ts are together

number of ways three positions can be taken out of the six vacant places = 6C3 = 20
hence, total number of ways = 120 x 20 = 2400 (option B)

shameekv1989 · Answer

In how many ways can letters the word ATTITUDE be rearranged such that no two Ts are adjacent to each other?

A. 1800
B. 2400
C. 3600
D. 4320
E. 6720

Total arrangements = 8!/3! = 6720 ways

Case I) All the T's are together - 6! = 720 ways

Case II) exactly 2 T's are together

For 1st position - 1 T out of 3 -> 3 ways
For 2nd position - 1 T out of the remaining 2 -> 2 ways
For 3rd position - No T -> 5 ways
For 4th position - Any remaining alphabets -> 5 ways
For 5th position - 4 ways
For 6th position - 3 ways
For 7th position - 2 ways
For 8th position - 1 ways

=> 3*2*5*5*4*3*2*1 = 3600

Therefore total combinations of no T's together = 6720-720-3600 = 2400 ways

Answer - B

Kinshook · Answer

Asked: In how many ways can letters the word ATTITUDE be rearranged such that no two Ts are adjacent to each other? A-1 T-3 I-1 U-1 D-1 E-1 Total words formed with letters of ATTITUDE = 8!/3! = 8*7!/6 = 4/3* 7! = 6720 Words in which 2 Ts are adjacent to one another = 7! = 5040 A-1 2T-1 T-1 I-1 U-1 D-1 E-1 But this includes the case when all 3 Ts are together and considers <2T>T & T<2T> as separate cases and are counted twice. Words in which 3 Ts are adjacent to one another = 6! = 120 Words in which one or more Ts are adjacent to one another = 7! - 6! = 6*6! = 4320 Words in which no Ts are adjacent to one another = 6720 - 4320 = 2400 IMO B

BrentGMATPrepNow · Answer

Take the task of seating the letters and break it into   stages  .

Stage 1 : Arrange the non-T letters (A, I, U, D and E) 
Since n unique objects can be arranged in n! ways, we can arrange these five letters in 5! ways. In other words, we can complete stage 1 in   120   ways.

Key step: For each arrangement of the 5 non-T letters, add a space on each side of each letter. So, for example, if we add spaces to the arrangement ADIUE, we get: _A_D_I_U_E_

Each of these six spaces represents a possible location for each of the 3 T's. Notice that this configuration ensures that no T's can be adjacent.

Stage 2 : Place T's in 3 of the 6 available spaces .
IMPORTANT: Since we're placing 3  identical  T's, the order in which select the 3 spaces doesn't matter. 
So, we can use combinations.
We can choose 3 of the 6 spaces in 6C3 ways (=   20   ways)
So, we can complete stage 2 in   20   ways.

At this point, we'll throw away the remaining spaces, leaving an arrangement with all 8 letters.

By the Fundamental Counting Principle (FCP), the number of ways to complete the 2 stages (and thus place all 7 letters) = (  120  )(  20  ) = 2,400 ways.

Answer: B

Cheers, 
Brent
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In how many ways can letters the word ATTITUDE be rearranged such that

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