How to solve this?

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Mon Apr 28, 2008 5:33 am
Thanked: 2 times

How to solve this?

by ramyaravindran » Mon Mar 08, 2010 5:51 am
If the letters of the word A, F, T, E, R are permuted and arranged in a dictionary form like after, then for some (n > 2) how many other possible ways are there?
A. 119
B. 117
C. 88
Source is TestMagic. I dont have the OA

Legendary Member
Posts: 610
Joined: Fri Jan 15, 2010 12:33 am
Thanked: 47 times
Followed by:2 members

by kstv » Mon Mar 08, 2010 8:08 am
The Q is not very clear. What is n ?

5 letters an be arranged in 5! ways = 120 ways of which AFTER is one of them so 120-1=119.

Senior | Next Rank: 100 Posts
Posts: 49
Joined: Mon Apr 28, 2008 5:33 am
Thanked: 2 times

by ramyaravindran » Mon Mar 08, 2010 10:35 am
I thought that they are asking for the number of all possible words that could appear after the word "AFTER" in the dictionary using letters A,F,T,E,R. The word should atleast be more than 2 characters in length. Let me know if this helps.

Senior | Next Rank: 100 Posts
Posts: 49
Joined: Mon Apr 28, 2008 5:33 am
Thanked: 2 times

by ramyaravindran » Mon Mar 08, 2010 1:54 pm
Can someone help with this question?

Senior | Next Rank: 100 Posts
Posts: 45
Joined: Tue Mar 02, 2010 4:27 am
Thanked: 2 times

by yeahdisk » Mon Mar 08, 2010 2:52 pm
This question is poorly constructed so I'm struggling to give you a proper answer.

As has already been said, you have 119 x 5 letter combinations (discounting AFTER)

You have 5C4 * 4! 4 letter combinations (120)
You have 5C3 * 3! 3 letter combinations (60)
You have 5C2 * 2! 2 letter combinations (20)

= 119 + 120 + 60 + 20 = 319

I think you need to confirm the question wording

Master | Next Rank: 500 Posts
Posts: 305
Joined: Mon Jul 27, 2009 5:38 am
Thanked: 10 times

by Shawshank » Tue Mar 09, 2010 1:47 am
Either there is something wrong with the answer choices or with the question
++++++++++++++++++++++++++++++
Shawshank Redemtion -- Hope is still alive ...

Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Fri Mar 12, 2010 2:51 am

by harsh001 » Sat Mar 13, 2010 4:08 am
Assuming n>2 meaning words with more than 2 alphabets need to be formed

Choosing 3: 5C3 * 3! = 60
Choosing 4: 5C4 * 4! = 120
Choosing 5 5C5 * 5! - 1(AFTER) = 119

Total = 60+120+119 = 299

I am not sure if i am right or not!!