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

Question

The letters A, D, R, O, I, and T can be used to form 6-letter strings as ADROIT or TDAROI. Using these letters, how many 6-letter strings can be formed which neither begin with T nor end in A ?

A. 6,480
B. 720
C. 528
D. 504
E. 480

How many words can be formed from the letters of the word ADROIT, which neither begin with T nor end in A ?

M38-01

Bunuel · Accepted Answer

GMAT Club Tests' Official Explanation:

The number of arrangement of six different letters A, D, R, O, I, and T,  without restrictions , is simply 6!.

What about the number of arrangements  with  the restrictions given?

(a) The number of arrangements which start with T is 5! (T is fixed as the first letter, T*****, and the remaining 5 letters can be arranged in 5! ways.).

(b) The number of arrangements which end with A is also 5! (A is fixed as the last letter, *****A, and the remaining 5 letters can be arranged in 5! ways.).

Now, (a) and (b) cases will have an overlap of arrangements which start with T AND end with A: T****A.

The number of such cases is 4! (T is fixed as the first letter, A is fixed as the last letter, T****A, and the remaining 4 letters can be arranged in 5! ways.).

So, the number of arrangements which start with T  OR  end with A is  5! + 5! - 4! .

The final answer is therefore  Total - Restriction = 6! - (5! + 5! - 4!) = 504 .

Answer: D.

Regor60 · Answer

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

Posted from my mobile device

Bunuel · Answer

______________________
Anyone else wants to try?

SIVAPRIYA1883 · Answer

total no of combinations without restrictions : 6!

no of words starting with T : 5!
no of words ending with A : 5!

for the condition neither T nor A ;

these 2 cases should be removed from the complete lot,

But the words starting with T and ending with A gets counted in both cases so getting removed twice,

so the words starting with T ending with A should be added once : 4!

so the final case turns out to be : 6!-5!-5! +4! = 504 .

happy to take any corrections

Regor60 · Answer

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

Posted from my mobile device

KarishmaB · Answer

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)

GMATWhizTeam · Answer

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.

Adarsh_24 · Answer

If someone wants to know why the normal dash method doesnt seem to work.

_ _ _ _

first letter can be A,D,R,O,I
Last letter can be D,R,O,I,T

So you may pick 5 _ _ _ _ 5

=> 5 * 4 * 3 * 2 * 1 * 5 =600 ( because both sides have used up 1 each, we start with 4.3.2.1 in remaining)
Here you are having combinations that start and end with same letter like D _ _ _ _ D.

Take the case for D where both first and last have fixed value.
1*4*3*2*1*1=24
Apart from T and A all other fours have this issue.
So 24*4=96

Ans=600-96=504

We have to be careful when repetition is not allowed.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=82768 Number of strings which neither begin with T nor end with A = Total number of possible strings (6!) - Hope this helps! ___ Harsha Passionate about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com GMAT-Club-Forum-cghdn4vu.png

bumpbot · Answer

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

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

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