The letters A, D, R, O, I, and T can be used to form 6-letter strings

Number of arrangements without restrictions:

6!= 720

With T in the first spot and A in any of the middle 4 spots, number of arrangements:

4(4!) = 96

A in the last spot and T in any of the middle 4 spots is the same as above, so another
96 arrangements.

Now subtract case where T is in first spot and A in last;

4!=24

Total arrangements:

720-192-24= 504

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______________________
Anyone else wants to try?

Yes. Missed the 24 beginning with T and ending with A.

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why do you add back 4!? how did you get there?

The string should not begin with T so there are 5 options for the starting letter.

Case 1: Starting letter is A.
We can make the string in 1*5*4*3*2 = 120 ways

Case 2: Starting letter is D/R/O/I
The first letter can be selected in 4 ways and the last letter also in 4 ways (except A and the letter chosen for first spot).
Now for the middle 4 spots, we have 4*3*2*1 = 24 arrangements.
Number of strings = 4*4*24

Total number of strings = 120 + 16*24 = 504 (only option with last digit 4)

Answer (D)

There are 6 distinct letters given: A, D, R, O, I and T.

The total number of 6-letter strings that can be made = \(6_P_6\) = 6! = 720.

Note that this represents ALL the 6-letter strings that can be made. However, we do not want ALL of them. We just need the strings which neither begin with T nor end with A. Therefore, answer options A and B can be eliminated.

Number of 6-letter strings which neither begin with T nor end in A = Total 6-letter strings – (Strings that begin with T) – (Strings that begin with A) + (Strings that begin with T and end with A).

Note that the last part is being added because you cannot subtract the same component twice.

6-letter strings that begin with T = \(5_P_5\) = 5! = 120. Since the strings begin with T, the other 5 letters can be arranged in 5 places in \(5_P_5\) ways.

6-letter strings end with A = \(5_P_5\) = 5! = 120. Since the strings end with A, the other 5 letters can be arranged in 5 places in \(5_P_5\) ways.

6-letter strings that begin with T and end with A = \(4_P_4\) = 4! = 24. Since the strings begin with T and end with A, the other 4 letters can be arranged in 4 places in \(4_P_4\) ways.

Therefore, required answer = 720 – 120 – 120 + 24 = 504.

The correct answer option is D.

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