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Q: a)How many odd three - digit numbers can be formed from

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Q: a)How many odd three - digit numbers can be formed from  [#permalink]

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New post 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?

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New post 16 Oct 2007, 09:19
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%
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New post 16 Oct 2007, 09:34
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.
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New post 16 Oct 2007, 10:18
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.
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New post 16 Oct 2007, 10:48
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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 3-digit odd numbers without repitition.

Simple .
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New post 16 Oct 2007, 11:22
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.
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New post 16 Oct 2007, 11:51
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.
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New post 16 Oct 2007, 12:54
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.
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New post 16 Oct 2007, 13:43
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 3-digit odd numbers without repitition.

Simple .
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New post 16 Oct 2007, 13:55
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.
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Re: Counting Problem  [#permalink]

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New post 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: 999-100+1=900

so 900*1/30=30
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Re: Counting Problem  [#permalink]

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New post 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]

200-100=100, so 100 numbers less than 200
999-100+1=900 total

100/900=1/9
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New post 16 Oct 2007, 14:54
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 3-digit 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
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New post 16 Oct 2007, 15:06
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?
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New post 16 Oct 2007, 15:18
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.
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New post 16 Oct 2007, 15:25
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.
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New post 16 Oct 2007, 18:01
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)
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New post 17 Oct 2007, 00:26
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 3-digit 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



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.

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Re: Q: a)How many odd three - digit numbers can be formed from  [#permalink]

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