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

1. 1800
2. 2400
3. 3600
4. 4320
5. 6720

“In how many ways can letters the word ATTITUDE be rearranged such that” - is a topic that is covered in the quantitative reasoning section of the GMAT. To successfully execute the GMAT Problem Solving questions, a student must possess a wide range of qualitative skills. The entire GMAT Quant section consists of 31 questions. The problem-solving section of the GMAT Quant topics requires the solution of calculative mathematical problems.

Solutions and Explanation

Approach Solution : 1

There are 5 additional distinct letters without the Ts, they are A I U D E
The number of ways these 5 letters can arrange themselves will be 5! which means 120.
There must be at least one letter between any two Ts in order to prevent them from being together. The arrangement can be written as _X_X_X_X_X_ , where each of the other letters is represented by an X.

This will prevent any two Ts from being together and allow the three Ts to fill any three of the six open positions.
The number of ways three positions can be taken out of the six vacant places = 6C3 = 20
Therefore the total number of ways = 120 x 20 = 2400

Correct Answer: (B)

Approach Solution : 2

A-1
T-3
I-1
U-1
D-1
E-1

Total words generated by the letters of ATTITUDE = (8!)/(3!) = (8*7!)/6 = 6720.
Words with 2 Ts next to each other equal 7! = 5040

A-1
2T-1
T-1
I-1
U-1
D-1
E-1

However, this also applies to the situation in which all three Ts are present and <2T>T & T<2T> are treated as separate cases and counted twice

Words with 3 Ts next to each other = 6! = 120
Words that contain one or more Ts next to each other = 7! - 6! = 6*6! = 4320
Words with no Ts being together = 6720 - 4320 = 2400

Correct Answer: (B)

Approach Solution : 3

Consider breaking the task of seating the letters into stages.

Stage 1: Group the letters that aren't Ts (A, I, U, D and E)
We can arrange these five letters in 5! different ways because there are n different ways to arrange n! unique objects.
In other words, there are 120 possible ways to finish stage 1.
It is important to add a space on either side of each letter in each arrangement of the five non-T letters. For instance, the arrangement for letters A,D,I,U,E becomes _A_D_I_U_E_ if spaces are added.
These six areas each represent a potential location for one of the three Ts. Notice how this arrangement prevents adjoining Ts from occurring.

Stage 2: Insert Ts into 3 of the 6 open spaces.
It is important to note that the order in which you choose the three spaces is irrelevant because we will be placing three identical Ts.
We can therefore use combinations.
We can select 3 of the 6 spaces in 6C3 ways
6C3= 20
We have 20 different ways to finish stage 2.
We will now remove the last few spaces, leaving a configuration with all 8 letters.
According to the Fundamental Counting Principle (FCP), the number ways to finish the two stages and place all seven letters will be 120*20 = 2,400 ways.

Correct Answer: (B)

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