The digits \(1\), \(2\), \(3\), \(4\), \(5\) are written down at random to form a five-digit number. Note; no digit may be repeated! Find: - how many such numbers are possible,
We have five ways to choose the first digit, and once we’ve done that, four ways of choosing the
second, and so on… So there are \(5! = 120\) possible numbers. - the chance that the last digit is odd,
The last digit must be one of \(1, 3\) and \(5\). Once we have chosen this, there are \(4!\) ways to choose the first four digits. Thus there are \(3 \times 4! = 72\) possible numbers that are odd and the chance that a number is odd is
\(\dfrac{72}{120}=\dfrac{3}{5}.\) Could we have just stated this anyway as \(3\) of our digits were odd? - how many of the numbers are divisible by \(4\).
Any five-digit positive integer can be written as \(100a + b\) for positive integers \(100 \le a < 1000\) and \(0 \le b < 100\). Then we only need to check \(b\)
for divisibility by \(4\) because \(100a + b = 4\left(25a + \frac{b}{4}\right)\) which is an integer if \(b\) is a multiple of \(4\). So the general rule for checking whether or not \(4\) goes into a number is this; does \(4\) go into the final two digits? If so, then the number is a multiple of \(4\), and not otherwise. (That’s how we check for leap years and the
next Summer Olympics.) The only pairs of digits that could end our number and which are divisible by \(4\) are \(12\), \(24\), \(32\) and \(52\). Thus, there are \(3! \times 4 = 24\) such numbers. There are two possible answers to this depending of if the number can start with a 0 or not. If it can then the number is $7 \times 6 \times 5 = 210$ However we don't usually write numbers with leading zeros. So assuming we don't allow a leading zero the answer is $6 \times 6 \times 5 = 180$
How many of these are odd? Lets consider picking the numbers in a special order First we need to pick a 1, 3 or 5 so in our first draw for the final
digit so we have a choice of 3. Next we pick our first digit which can be any thing apart from 0 or the number we have just picked making a choice of 5 and for our final draw the middle digit we can pick any of the 5 remaining digits making $3 \times 5 \times 5 = 75$
Finally For the third one greater than 330 we have two ways to achieve this either draw a 4, 5 or 6 for the first digit then we don't care about the others making $3 \times 6 \times 5 = 90$ Alternatively we must draw
a 3 for our first digit then 4, 5 or 6 for our second and finally any digit making $1 \times 3 \times 5 = 15$ Now since both of these groups have no overlap we can simply add them together to get $90 + 15 = 105$.
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Intern Joined: 09 Aug 2006 Posts: 11 Q: a)How many odd three - digit numbers can be formed from [#permalink]
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? Manager Joined: 02 Aug 2007 Posts: 54 Re: Q: a)How many odd
three - digit numbers can be formed from [#permalink] 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% Intern Joined: 09 Aug 2006 Posts: 11 Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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. Intern Joined: 09 Aug 2006 Posts: 11 Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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. Intern Joined: 09 Aug 2006 Posts: 11 Re: Q: a)How many odd three - digit numbers can be formed
from [#permalink]
16 Oct 2007, 10:48 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 .
Intern Joined: 09 Aug 2006 Posts: 11 Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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. Intern Joined:
09 Aug 2006 Posts: 11 Re: Q: a)How many odd three - digit numbers can be formed from [#permalink]
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. Intern Joined: 27 Sep 2007 Posts:
10
Re: Q: a)How many odd three - digit numbers can be formed from [#permalink]
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. Intern Joined: 16 Oct 2007 Posts: 10 Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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 . Intern Joined: 27 Sep 2007 Posts: 10 Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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. Director
Joined: 09 Jul 2007 Posts: 612 Location: London Re: Q: a)How many odd three - digit numbers can be formed from [#permalink]
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 Director Joined: 09 Jul 2007 Posts: 612 Location: London Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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 Manager Joined: 07 Sep 2007 Posts: 70 Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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
Intern Joined: 27 Sep 2007 Posts: 10 Re: Q: a)How many odd three - digit numbers can be formed from [#permalink]
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? Intern Joined: 09 Aug 2006 Posts: 11 Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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. Intern Joined: 27 Sep 2007 Posts: 10 Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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. Manager Joined: 07 Sep 2007 Posts: 70 Re: Q: a)How many odd three - digit numbers can be formed from
[#permalink]
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) Senior Manager Joined: 09 Aug 2006 Posts: 394 Q: a)How many odd three - digit numbers can be formed from [#permalink]
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. Non-Human User Joined: 09 Sep 2013 Posts: 24597 Re: Q: a)How many odd three - digit numbers can be
formed from [#permalink] 24 Dec 2018, 14:26 Hello from the GMAT Club BumpBot!
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_________________ Re: Q: a)How many odd three - digit numbers can be formed from [#permalink]
24 Dec 2018, 14:26
How many four
And the hundred's also the same, 8 possible numbers, and coming to the ten's place, we will have only 7 possible numbers to place in it. So the number of four-digit odd numbers can be formed is 2240.
Hence, the number of 3-digit odd numbers that can be formed =3×6×6=108.
Hence, the required number of numbers =504.
As stated in the title above: How many 4-digit odd numbers can be formed using the digits 0, 1, 2 and 3 only if the repetition of the digit is not allowed? I already have the answer for this and it is 8.
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