M38-01

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

Bunuel · Accepted Answer

Official Solution:

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

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

An78w · Answer

Hi Banuel,

The way I solved this was -

6 positions for letters -

Position 1 - 5 options (except T)
Position 6 - 4 options (except A and the letter placed in position 1) 
Then 4! for rest 4 positions.

4! x 5 x 4 = 480.

Conceptually what am I missing?

Bunuel · Answer

X X X X X X

If A is the first letter, then the sixths letter has 5 choices ( NOT 4, as in your solution ), the fifths has 4 choices, the fourths has 3 choices, the third has 2 choices and the second has 1 choice. So, if A is the first letter then we have 1*5*4*3*2*1 = 120.

If the first letter is D, R, O, or I (4 choices), then the sixths letter has 4 choices, the fifths has 4 choices, the fourths has 3 choices, the third has 2 choices and the second has 1 choice. So, if the first letter is D, R, O, or I, then we have 4*4*4*3*2*1=384. ( This case follows your logic )

Total = 120 + 384 = 504.

As you cane see, if A is the first letter then the sixths letter has 5 choices, not 4 as in your solution! So, you should consider those two scenarios (A is the fits letter/A is NOT the first letter) separately to get the correct answer.

The question was built with this trap in it, hence 480 as an option!

Hope it's clear and you liked the question!

VeeB · Answer

When we’re putting T in the first place and A in the last place by fixing them, we are only making grouping of 6 where 
where T is always in the first place and A in the last place.
Therefore the 6 letter grouping where T is not in the first place and A not in the last place should be 6!-4!
Where’s my logic going strayed?

Posted from my mobile device

Bunuel · Answer

This is explained in detail in the solution, but here you go again:

You are excluding the case where T is the first letter and A is the last letter (T****A). However, you should also exclude the cases where T is the first letter and A is any other letter (T*****), and where A is the last letter and T is any other letter (*****A).

WarasSingh · Answer

It is not explicitly mentioned that repetition is not allowed. How do we infer that? or does a string mean set of different number?

Bunuel · Answer

From the wording, it should be clear that only the letters A, D, R, O, I, and T are used, each exactly once. You can also check this official question with the same setup for reference: https://gmatclub.com/forum/the-letters- ... 20320.html Hope it helps.

QuantMadeEasy · Answer

I have solved like this:

https://www.youtube.com/watch?v=GC0AmZa5EuA

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SudsMeister · Answer

Remembering the 2-set overlapping formula helped me in solving this effectively.

Question asks for neither begin with T, nor end in A.

In 2-set overlap, Total items = (items in set A) + (items in set B) - (items in set A and B) + (items neither in Set A, nor in Set B)
Here, Total items = Number of possible 6-letter strings
Set A = strings that begin with T
Set B = strings that end in A
Items in Set A and B = strings that begin with T and end in A
Items neither in Set A, nor in Set B = what we have to find.

Rewriting it in shorthand, we get, T = A + B - AB + N
N = T - A - B + AB

T = Arranging 6 letters to make 6-letter strings = 6! = 720

A = strings that begin with T = T is fixed in the first spot, remaining 5 spots can be taken by remaining 5 letters in 5! ways = 120

B = strings that end in A = A is fixed in the last spot, remaining 5 spots can be taken by remaining 5 letters in 5! ways = 120

AB = strings that begin with T and end in A = T is fixed in the first spot, A is fixed in the last spot, remaining 4 spots can be taken by remaining 4 letters in 4! ways = 24

N = 720 - 120 - 120 + 24 = 504.

Answer D.

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