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mettlesomedinesh
Associate
Joined: 29 Mar 2008
Posts: 1
Location: Chennai
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 Posted: Thu Apr 10, 2008 11:49 am Post subject: Permutation and Probablity Questions |
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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. |
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ksb
GMAT Tutor
Joined: 28 Dec 2004
Posts: 54
Location: Chennai, India
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| Posted: Wed Apr 23, 2008 5:19 pm Post subject: Re: Permutation and Probablity Questions |
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| mettlesomedinesh wrote: |
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! ) ? |
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.
| mettlesomedinesh wrote: |
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!. |
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.
| mettlesomedinesh wrote: |
3) A bag 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. |
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} |
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