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These are questions from permuation and combination topics, from rearraging words sub topic.

Can some one please help me in this. I have got answer, but not sure of whether it is correct.

1) In the word "management" , last 3 letters "ent" is fixed in position In how many ways will the remaining letters be rearranged.
Is it 7! / ( 2! * 2! ) ?

2) In the word "MANAGEMENT" AAEE, In how many possible ways AAEE appear together?
Is it 7! / ( 2! * 2! ) ?.. This is obtained by having AAEE together and permuting others. since in the word aaee repeats i have divided by 2! and 2!.

3) A bang contains 5 red balls, 7 green balls, 6 white balls. When 3 balls are drawn at random from this bag, what is the probablity that 2 balls are of same color and one is different color.

Yes the answer is correct. The first 7 letters can be rearranged in 7! (2! * 2!) ways. One of the 2! for the two As and the other one for the 2 Ms.



The answer is correct. However, the division by the two 2!s is for the two Ms and the two Ns and not for the two As and Es.

The question is ambiguous as it does not state if the order in which AAEE appear the same way when they appear together or one can rearrange those 4 letters, keeping them together as a group.



Two balls belong to same color and one to a different color could be as follows

1. 2 red, 1 green
2. 2 green, 1 red
3. 2 green, 1 white
4. 2 white, 1 green
5. 2 red, 1 white
6. 2 white, 1 red

So, required probability is {5C2*7C1 + 5C1*7C2 + 7C2*6C1 + 6C2*7C1 + 5C2*6C1 + 6C2*5C1} / 18C3

Actually, when 3 balls are taken, either all are same color or all or different colors or two belong to a color and the third one a different color.

So, the answer can also be written as

1 - {probability that all belong to same color + probability that all belong to different color}

1 - { (5C3 + 6C3 + 7C3)/18C3 + (5C1*6C1*7C1)/18C3}