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Intern
Joined: 09 Aug 2006
Posts: 13

Q: a)How many odd three  digit numbers can be formed from
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16 Oct 2007, 08:49
Q:
a)How many odd three  digit numbers can be formed from the digits (0,1,2,3,4,5) if each digit is to be used only once?
b) If we randomly pick one from these numbers , what is the probability that it is less than 200? == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



Manager
Joined: 02 Aug 2007
Posts: 142

This is a combinations question:
a)
Digit 3  5 possibilities (0 cannot be used in this digit)
Digit 2  6 ''
Digit 1  6 ''
Total 3 digit possibilities = 5*6*6 = 180
Odd 3 digit possibilities = 180 / 2 = 90 numbers
b) Since digit 3 is 2, we only need to find the total possibilities of the final two digits = 6*6 = 36.
So Expected outcome / total outcomes = 36/90 = 2/5 = .40 = 40%



Intern
Joined: 09 Aug 2006
Posts: 13

The Question says that how many "ODD" numbers can be formed. So "Digit number 3" should not have 5 possibilities, as well as Digit number 1 and 2 should not have 6 possibilities.
Try again.



Intern
Joined: 09 Aug 2006
Posts: 13

Nope. Wrong. The answer which i have is different from the one which you people are giving.
Olga, why do you think that the first digit should not be zero. The digits can be like 015, 021 etc etc.



Intern
Joined: 09 Aug 2006
Posts: 13

Well, i explain the answer of part a.
The problem says that we have to choose 3 odd  digit numbers from 0,1,2,3,4,5 if each digit can only be used once.
The last digit can be chosen in 3 ways(1,3,5) so we have 3 choices for the last digit.
First digit can be chosen in 4 ways assuming that 0 can't be chosen as the first digit and the last digit has already been chosen.
Second digit can also be chosen in 4 ways assuming that the last and first digits have been already chosen.
By the product rule 4.4.3 = 48 ways to choose 3digit odd numbers without repitition.
Simple .



Intern
Joined: 09 Aug 2006
Posts: 13

b) If you read the question properly, it says that if we pick a number randomly from the numbers we have already got from part a, what is the probability that it is less then 200?
So it can't contain all the numbers between 100 to 200 , as we can't count 107, 109 etc , as in part a, we form odd numbers from 0,1,2,3,4,5.
Therefore considering this, we have only 1 choice for the 1st digit.
As the numbers shouldn't repeat , so we have only 2 choices left for the last digit, b/c we already have 1 in the first digit.
Once the 1st and last digit are chosen , we have 4 choices for the 2nd digit.
So by product rule we have 1.2.4 = 8 numbers that are less then 200 and contain only the digits (0,1,2,3,4,5).
So , P = 8 /48 = 1/6.
And that is also the official answer.



Intern
Joined: 09 Aug 2006
Posts: 13

First digit can't be chosen in 5 ways. It should be in 4 ways because we only have 4 choices left after the 0 is omitted AND after the last digit has been chosen. So we should omit "0" and the "last digit , whatever it is" from our choices.
Same is the case with the second digit.



Intern
Joined: 27 Sep 2007
Posts: 16

Quote: First digit can't be chosen in 5 ways. It should be in 4 ways because we only have 4 choices left after the 0 is omitted AND after the last digit has been chosen. So we should omit "0" and the "last digit , whatever it is" from our choices.
Same is the case with the second digit.
you are wrong here.
Look: when you exclude 0 from the first digit, one digit numbers are also excluded cause 1 = 01 = 001 (u have excluded 2 digit numbers and in this situation one digit numbers appear to be 2 digit numbers indeed with 0 leading), so first digit you can choose in 5 ways and the next one to the end you can choose in 6 ways.



Intern
Joined: 16 Oct 2007
Posts: 22

Well AlexTsipkis:
I get a slightly diff answer I dont know what is wrong in my approach.
I start from the last digit and move 3,2,1
last digit should be odd hence (1,3,5) = 3 ways
2nd digit no restriction but 1 already gone = 5 ways
1st digit cannot be zero + 2 already gone = 3 ways
3*5*3 = 45 ways!
Thx
AlexTsipkis wrote: Well, i explain the answer of part a.
The problem says that we have to choose 3 odd  digit numbers from 0,1,2,3,4,5 if each digit can only be used once.
The last digit can be chosen in 3 ways(1,3,5) so we have 3 choices for the last digit.
First digit can be chosen in 4 ways assuming that 0 can't be chosen as the first digit and the last digit has already been chosen.
Second digit can also be chosen in 4 ways assuming that the last and first digits have been already chosen.
By the product rule 4.4.3 = 48 ways to choose 3digit odd numbers without repitition.
Simple .



Intern
Joined: 27 Sep 2007
Posts: 16

Quote: Quote: First digit can't be chosen in 5 ways. It should be in 4 ways because we only have 4 choices left after the 0 is omitted AND after the last digit has been chosen. So we should omit "0" and the "last digit , whatever it is" from our choices.
Same is the case with the second digit.
you are wrong here. Look: when you exclude 0 from the first digit, one digit numbers are also excluded cause 1 = 01 = 001 (u have excluded 2 digit numbers and in this situation one digit numbers appear to be 2 digit numbers indeed with 0 leading), so first digit you can choose in 5 ways and the next one to the end you can choose in 6 ways.
i didn't see that that each digit should be used only once
so it semms to me that ramubhaiya's approach is right.



VP
Joined: 09 Jul 2007
Posts: 1054
Location: London

Re: Counting Problem
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16 Oct 2007, 14:45
a)How many odd three  digit numbers can be formed from the digits (0,1,2,3,4,5) if each digit is to be used only once?
6 digits to choose from
xx(1or3or5)
last digit: 3/6
second digit: 1/5 (any digit other than the last one thus one out of five)
first digit: 1/3 (0 out, second and the last digits chosen already, so 1 out of 3 left)
thus
(1/3)*(1/5)*(3/6)=3/90=1/30
now how many three digit numbers are there: 999100+1=900
so 900*1/30=30



VP
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Location: London

Re: Counting Problem
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16 Oct 2007, 14:47
b) If we randomly pick one from these numbers , what is the probability that it is less than 200?[/quote]
200100=100, so 100 numbers less than 200
999100+1=900 total
100/900=1/9



Manager
Joined: 07 Sep 2007
Posts: 108

ramubhaiya wrote: Well AlexTsipkis: I get a slightly diff answer I dont know what is wrong in my approach. I start from the last digit and move 3,2,1 last digit should be odd hence (1,3,5) = 3 ways 2nd digit no restriction but 1 already gone = 5 ways 1st digit cannot be zero + 2 already gone = 3 ways 3*5*3 = 45 ways! Thx AlexTsipkis wrote: Well, i explain the answer of part a.
The problem says that we have to choose 3 odd  digit numbers from 0,1,2,3,4,5 if each digit can only be used once.
The last digit can be chosen in 3 ways(1,3,5) so we have 3 choices for the last digit.
First digit can be chosen in 4 ways assuming that 0 can't be chosen as the first digit and the last digit has already been chosen.
Second digit can also be chosen in 4 ways assuming that the last and first digits have been already chosen.
By the product rule 4.4.3 = 48 ways to choose 3digit odd numbers without repitition.
Simple .
You have to choose the 100s digit before the 10s digit. If you choose the 10s digit, you no longer have symmetry in your outcomes. (Choosing '0' for the 10s digit allows 4 choices where choosing any other number allows 3 choices.)
If you wanted to do it your way:
3 ways to choose 1s digit
5 ways to choose 10s digit
(1/5)*4 + (4/5)*3 ways to choose 100s digit
3 * 5 * ((1/5)*4 + (4/5)*3) = 48 ways



Intern
Joined: 27 Sep 2007
Posts: 16

Quote: (1/5)*4 + (4/5)*3 ways to choose 100s digit
plz, could you explain this expression ((1/5)*4 + (4/5)*3), how did you recive that, i mean theory?



Intern
Joined: 09 Aug 2006
Posts: 13

Djames,
Your approach and understanding of the problem is totally Wrong.
I have explained my approach and my approach and understanding gives me the right answer which is the OA.



Intern
Joined: 27 Sep 2007
Posts: 16

AlexTsipkis
i agree with you, i was wrong and i've got your explanation.
Now i'm interested in JingChan approach, cause i want to understand what he did.



Manager
Joined: 07 Sep 2007
Posts: 108

Djames wrote: AlexTsipkis i agree with you, i was wrong and i've got your explanation. Now i'm interested in JingChan approach, cause i want to understand what he did.
Don't get me wrong, I used Alex's approach when I did the problem.
For the 10s choice:
1 choice (0) allows you to get 4 choices for the 100s
::> 4
4 choices allows you to get 3 choices for the 100s
::> 3+3+3+3
So you take the 3 for the 1s digit multiplied by the number of outcomes from the remaining digits.
::> 3 * (4+3+3+3+3) = 48
This would be the same as:
::> 3 * 5 * avg(choices for 100s place)



Director
Joined: 09 Aug 2006
Posts: 730

JingChan wrote: ramubhaiya wrote: Well AlexTsipkis: I get a slightly diff answer I dont know what is wrong in my approach. I start from the last digit and move 3,2,1 last digit should be odd hence (1,3,5) = 3 ways 2nd digit no restriction but 1 already gone = 5 ways 1st digit cannot be zero + 2 already gone = 3 ways 3*5*3 = 45 ways! Thx AlexTsipkis wrote: Well, i explain the answer of part a.
The problem says that we have to choose 3 odd  digit numbers from 0,1,2,3,4,5 if each digit can only be used once.
The last digit can be chosen in 3 ways(1,3,5) so we have 3 choices for the last digit.
First digit can be chosen in 4 ways assuming that 0 can't be chosen as the first digit and the last digit has already been chosen.
Second digit can also be chosen in 4 ways assuming that the last and first digits have been already chosen.
By the product rule 4.4.3 = 48 ways to choose 3digit odd numbers without repitition.
Simple . You have to choose the 100s digit before the 10s digit. If you choose the 10s digit, you no longer have symmetry in your outcomes. (Choosing '0' for the 10s digit allows 4 choices where choosing any other number allows 3 choices.) If you wanted to do it your way: 3 ways to choose 1s digit 5 ways to choose 10s digit (1/5)*4 + (4/5)*3 ways to choose 100s digit3 * 5 * ((1/5)*4 + (4/5)*3) = 48 ways
Jingchan, Can you pls. explain what you mean by 'symmetry in your outcomes'? Is this always the case? For example what if this was a 4 digit odd integer that we were choosing?
Also can you pls. explain how you get (1/5)*4 + (4/5)*3 ways to choose 100s digit?? thanks. == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: Q: a)How many odd three  digit numbers can be formed from
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24 Dec 2018, 14:26
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Re: Q: a)How many odd three  digit numbers can be formed from &nbs
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24 Dec 2018, 14:26






