How many 5 letter combinations can be made from the letters of the wor

D should be the answer.

1st Place: Vowel: 2C1=2
2-4 Place: 5*4*3=60
Last Place: 5C1= 5

Multiply all of them: 600

IMO the answer is option D

we have 5 places: _ _ _ _ _
first place has 2 options , last place has 5 options now, we have 5 options left for 3 places so, 2 x 5 x4 x 3 x 5 = 600

I just wanted to go though my mistake real quick, for those who may have picked the trap answer.

I made the problem more complicated than it had to be. I tried breaking the counting down into two cases:

Case 1: 2 vowels are picked.


Case 2: 1 vowel is picked


Then, for each case, I would place a vowel in the first letter and a consonant in the last letter. Then, I would try to determine the different arrangements ———> over counting ———> 720 arrangements


The proper way to do this kind of Arrangement problem in which conditions are placed on certain spaces is to use the “slot method” or “box and fill method” or whichever fancy title that may have been used.


_____; ____ ; _____ ; ____ ; ____
Slot 1 ; slot 2 ; slot 3 ; slot 4 ; slot 5


We have:

5 available consonants (V , R , M , N, T)
And
2 available vowels (E , O)

Slot 1 and Slot 5 are both constrained. Using this type of method, one should always start with the most constrained “slots” first


Slot 1: since a vowel must be chosen, we have 2 available options from which to choose —— “2 choose 1” = 2

Slot 5: since a consonant must be chosen, we have 5 available options from which to choose ——— “5 choose 1” = 5

Assume E was chosen for slot 1 and V was chosen for slot 5

We have 4 consonants and 1 vowel —-> 5 letters from which to choose and 3 available spots left

Step 1: how many different combinations of 3 letters can we choose to fill those 3 spots

“5 choose 3” = 5! /3! 2! = 10

AND

Step 2: for each possible combination of 3 letters, how many different ways can we arrange those 3 letters in the 3 spots left

3! = 6

Total Count of Unique Arrangements:

(2) * [ 10 * 6 ] * (5) =

600 unique arrangements

Posted from my mobile device

Word: VERMONT
Total letters: 7

Vowels: 2 = E and O
Consonants: 5 = V,R,M,N,T

Condition: 5 letter combination

First letter vowel: ^2{C_1} = 2
Last letter consonant: ^5{C_1} = 5

The remaining 3 places will have 5 * 4 * 3 options = 60

Total: 2 * 5 * 60 = 600

Answer D

How many 5 letter combinations can be made from the letters of the word VERMONT if the first letter has to be a vowel and the last letter has to be a consonant, and each letter can be used only once?

For 1st place, there are two vowels = 2C1 = 2 ways
for last place, there are 5 consonants = 5C1 = 5 ways

For remaining 3 places = 5C3 *3! = 60

So, total ways = 2*5*60 =600. I think D.

First letter can be chosen in 2(E,O) ways.
Last letter can be chosen in 5(VRMNT) ways.
There are 7 alphabets and the required word has 5 alphabets.
_ _ _ _ _
2 _ _ _ 5

The remaining three positions can be filled in 5*4*3 ways i.e. \(5_{C_3}\) ways.
Ways in which the 5 letter word can be formed = 2*5*4*3*5 = 600

Answer D.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.