combination Q

[gmatfordays]

In how many ways can the letters of the word MANAGEMENT be rearranged so that the two As do not appear together?

(A) 10! - 2!
(B) 9! - 2!
(C) 10! - 9!
(D) None of these

[Anony]

Is the answer=(1/16)*(10!-2*9!)??

[rumen]

the total different perm of MANAGEMENT are 10!/2!2!2!2!. When 2 A's are together then we have 9!/2!2!2!2!.This is equivalent to 10!-9! or in 10 arrangements A's are separated.

[kpadma]

Originally posted by rumen

the total different perm of MANAGEMENT are 10!/2!2!2!2!. When 2 A's are together then we have 9!/2!2!2!2!.This is equivalent to 10!-9! or in 10 arrangements A's are separated.

the total different perm of MANAGEMENT are 10!/2!2!2!2!. When 2 A's are together then we have 9!/2!2!2!2!.
I understand upto this point.

This is equivalent to 10!-9! or in 10 arrangements A's are separated.
I don't understand? Could you elaborate on this?
:o

[dig12us]

My answer is D. None of these.

Here is what I did, Please let me know where do you think am I wrong

Ways in which 2 A's are separated = Total Ways - Number of ways they appear together

Total ways = 10!

Number of ways they appear together = 9! 2!

so Ans is 10! - 9!(A's together so 9 units) 2! (2As can be arrenged in 2! ways within themselves)

[rumen]

You are right not to understand since I made a silly mistake, my bad.:oThe difference is NOT 10 but another number.Since 2 A's are one element we have one 2! less in denominator or 9!/2!2!2!. When you substract 9!/2!2!2! from 10!/2!2!2!2! you will get the arrangements without 2A's together.10!/2!2!2!2!-9!/2!2!2!, multiply second number denominator and nominator with 2! so that we can substract it and we have10.9!-2.9!/2!2!2!2!=8.9!/2!2!2!2!=9!/2! Also think that the answer should be d.
Dig12us , agree with your reasoning how to find the perm without 2A's , but please note that the DIFFERENT permutations are 10!/2!2!2!2! since we have repetition of the letters A,M,N,E. Why do you multiply 9! with 2!? You have 2 A's, with the two arrangements- AA, AA .Can you say if they are different arrangements?I do not think so. Additionally, we count 2A's as one element.

[dig12us]

rumes

YES you are right. my misatke

[Anony]

Dear Rumen!
Mine first response was exactly what you have explained, though a bit abstract, but thanx for the explaination.

[gmatfordays]

Hi all,

D is the correct answer. Here's how:

The word MANAGEMENT is a 10 letter word.

Normally, any 10 letter word can be rearranged in 10! ways.

However, as there are certain letters of the word repeating, we need to account for those. In this case, the letters A, M, E and N repeat twice each.

Therefore, the number of ways in which the letters of the word MANAGEMENT can be rearranged reduces to . 10!/2!x2!x2!X2!

The problem requires us to find out the number of outcomes in which the two As do not appear together.

The number of outcomes in which the two As appear together can be found out by considering the two As as one single letter. Therefore, there will now be only 9 letters of which three of them E, N and M repeat twice. So these 9 letters with 3 of them repeating twice can be rearranged in 9!/2!x2!x2! ways.

Therefore, the required answer in which the two As do not appear next to each other =

Total number of outcomes - the number of outcomes in which the 2 As appear together

=> 10!/2!x2!x2!x2! - 9!/2!x2!x2! ways.