In how many ways can you arrange the letters "APTITUDE" so as to form

Question

In how many ways can you arrange the letters "APTITUDE" so as to form words that begin and end with a vowel?

A. 640
B. 1280
C. 4320
D. 8640
E. 9860

Bunuel · Accepted Answer

We have 4 vowels, as order matters in this case # of arrangements of 4 vowels in 2 slots (begin/end) will be 4P2=12

Now we are left with 8-2=6 letters, from which 2 letters are the same (T). # of arrangements 6!/2!=360

Total=360*12=4320

Answer: C.

So I think the answer is right.You just didn't consider that order matters in the first case and repeated T in the second case, but these two mistakes compensated each other and you got the correct answer.

Hope it's clear.

prome2 · Answer

APTITUDE has 8 letters, so given that the letters start and end with a vowel there are 2 conditions.

1) There are 4 choose 2 possibilities for a vowel start/end combination.

2) The remaining 6 letters can be arranged in any order and still form a word. Thus permutate 6 objects.

Answer = (4!/2!2!)*6!= 4320

Edit : Given T is repeated and non vowel. I think the final answer should be divided by 2! to give 2160 (which is not an option)

Would love to get some advise!

prome2 · Answer

Crystal! Thanks Bunuel! +1

chetan2u · Answer

Hi...

The first letter of the word can be ANY of the 4 vowels and the last letter could be any of the remaining 3..
So first and last letter can be selected in 4*3 ways.
So ans has to be a MULTIPLE of 3... Only C and D are left ( just for info)

Now remaining 6 can be selected in 6!/2!=360... Div by 2 is because of 2 I..

Ans 360*12=4320

C

shashankism · Answer

There are 4 vowels AIEU
SO 1st letter can be taken in 4 ways
Last letter can be taken in 3 ways
Middle letters can be arranged in 6! ways from 6 letters left.

So total ways = 4*3*6!

But there are 2 T's so we need to divide total arrangement by 2! to compensate the duplicate of words.

Total required no. words formed = 4*3*6!/2! = 4320

Answer C

ScottTargetTestPrep · Answer

If the first letter has to be a vowel, then it can be any one of the 4 vowels (A, E, I, U). If the last letter also has to be a vowel, then it can be any one of the remaining 3 vowels. After we have selected the two vowels for the first and last positions, we have 6 letters left, of which 2 repeat. Thus, the number of ways to arrange these in-between letters is 6!/2! = 720/2 = 360. Finally, the number ways the words can be formed is 4 x 360 x 3 = 4,320.

Answer: C

mvmarcus7 · Answer

The way I did was a bit longer than some answers, but helped me understand better the problem.

First I thought there are 8 letters in total and marked on the paper 8 spaces for the counting. The first letters will be able to use 4 potential letters, and the last one, 3 remaining ones, since there are 4 vowels.

Then, for the second space, I got 8 -2 (total - 2 vowels used) = 6, for the 3rd space I got 5 and so on. After multiplying it all, I got (120 x 72), but, since T is counted twice, I divided by 2 to remove double counting, which resulted in 60 x 72 = 4,320

CEdward · Answer

Permutation with restrictions.

V _ _ _ _ _ _ V

4 <--- # of ways a vowel can be chosen for the first letter
3 <--- # of ways... last letter

4 x 3 = 12 possibilities

6 letters left...

6! / 2! = 360 <--- # of ways the remaining letters (including two vowels can be arranged)

360 x 12 = 4320 <--- # of possible words with vowels in first/last position

Answer is C.

testtakerstrategy · Answer

Question -- why doesnt this method work:

8!/2! total ways to arrange the letters
A. ½ of the total will have a VOWEL as the first letter (since ½ of the total letters are vowels).
B. Similarly, ½ of the total will have a VOWEL as the last letter (since ½ of the total letters are vowels).

therefore we can take the total number and divide by 4-> ¼ of all will have a vowel at the beginning and end:

(8!)/(2!4)

Why is this flawed?

juanar · Answer

This question is tricky because, for example ATIUDETP is not a valid word and nor are many of the 4320 possible arrangements

officiisnobis · Answer

Number of ways for first letter - 4
Number of ways for last letter - 3
Number of ways for others ( between last and first) - 6!/2!
Total - 4* 3 * (6!/2!) = 4,320

In how many ways can you arrange the letters "APTITUDE" so as to form

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In how many ways can you arrange the letters "APTITUDE" so as to form

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards