Latest CTS Aptitude Question SOLUTION: In how many ways the letters in the word MANAGEMENT can be arranged such that A’s does not come together.

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In how many ways the letters in the word MANAGEMENT can be arranged such that A’s does not come together.

Solution

Methods to arrange all the letters leaving A = 8*7*6*5*4*3*2*1/(2*1)*(2*1)*(2*1)= 5040
Now between these 8 letters , there are 9 places so methods to place A and A in these positions = 9*8/2*1 = 36
thus total methods are 5040*36 = 181440

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Asked In CTS Anil (13 years ago)

Solved Read Solution (3)

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Read Solution (3) :

Shobin Thomas  (13 years ago)

Part 1:
The word MANAGEMENT can be arranged in 10! ways if we don't take any constraints.

Part 2:
Now consider the case of arranging the letters of the word MANAGEMENT such that, in all its arrangements, both A's will appear together. Like: MNGEMENT(AA), M(AA)NGEMENT, (AA)MNGEMENT, etc. Assume both A's to be one single character $. So it would be interpreted something like MNGEMENT$, M$NGEMENT, $MNGEMENT, etc. These set of 9 characters, can be arranged in 9! ways.

Part 3:
Now subtracting both the values of "All possible arrangements" and "Arrangements of both A's together" would give

Is this solution Helpfull? Yes (9) | No (15)  

Aniket Sanjay Deshmukh  (9 years ago)

there are total 10 letters..
no. of M = 2
no. of A = 2
no. of N = 2
no. of E = 2
and others appear only one...
no. of ways to arrange these 10 letters = 10! / [ 2! * 2! * 2! * 2!]
= 226800
now, assume that two A i.e AA as single letter, total 9 letters
no. of ways to arrange these 9 letters where two 'A' always together
= 9! / [ 2! * 2! * 2! ]
= 45360

no. of ways of arrangement where two 'A' do not come together = 226800-45360
= 181440

Is this solution Helpfull? Yes (9) | No   

Rishav Kumar  (8 years ago)

since, there are 8 letters except A, hence we can place A in 9 different position
so, the no. of ways = 9p2 * 8p8 / 2p2 * 2p2 * 2p2 * 2p2 = 181440

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