$\begingroup$
We can think to this "scheme": you have five ordered empty "boxes" and each of them has to be filled with one digit from $0,\ldots,5$ [which are $6$ "objects"] with no repetitions, in order to get a $5$-digits number. This means that the last two digits must form a number which is divisible by $4$; then the possible last two digits are as follows: $04$, $12$, $20$, $24$, $32$, $40$, $52$. Now, take for instance $04$; how many numbers [strings] of five distinct digits ending with $04$ are there? Since the last two digits are taken, we have to choose the first three among the remaining ones, that is we have to choose and order $3$ objects out of the $4$ left: this means we have $D_{4,3} = 4 \cdot 3 \cdot 2$ such strings. The same reasoning applies to $20$ and $40$, but not to $12$, $24$, $32$ and $52$: in these cases, in facts, for the remaining three digits $0$ is available but cannot be chosen as first digit, since we want $5$-digits numbers. Thus, in these four cases, the first digit can be filled in $3$ ways [and not $4$, because $0$ is also excluded], the second in $3$ ways and the third in $2$, giving $3\cdot 3\cdot 2$ possible numbers for each couple of final two non-zero digits. Hence, since for $04$, $20$ and $40$ [which are $3$ "cases"] we apply the first reasoning, and for $12$, $24$, $32$ and $52$ [$4$ "cases"] the second one, the total amount of numbers of five distinct digits [chosen among $0,\ldots,5$] that are divisible by $4$ is $$3 \cdot [4 \cdot 3 \cdot 2] + 4 \cdot [3 \cdot 3 \cdot 2] = 144.$$
answered May 20, 2017 at 10:19
$\endgroup$
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Hello Guest!
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join 700,000+ members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan Prep. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application status, and more
Books/Downloads
Download thousands of study notes, question collections, GMAT Club’s Grammar and Math books. All are free!
and many more benefits!
- Register now! It`s easy!
- Already registered? Sign in!
Sep 15
Now through September 20th, get 20% off GMAT courses, practice exams, question banks, and more with Manhattan Prep Powered by Kaplan.
Sep 24
Does GMAT RC seem like an uphill battle? e-GMAT is conducting a masterclass to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days.
Sep 25
Attend this webinar to learn the core NP concepts and a structured approach to solve 700+ Number Properties questions in less than 2 minutes.
Manager
Joined: 29 May 2008
Posts: 91
How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
[#permalink]
00:00
Question Stats:
Hide Show timer Statistics
How many five digit numbers can be formed using digits 0, 1, 2, 3, 4, 5, which are divisible by 3, without any of the digits repeating?
A. 15
B. 96
C. 120
D. 181
E. 216
Originally posted by TheRob on 22 Oct 2009, 13:20.
Last edited by Bunuel on 10 Jun 2021, 06:07, edited 2
times in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 86798
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?
A. 15
B. 96
C. 120
D. 181
E. 216
First step:
We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.
We have six digits: 0,1,2,3,4,5. Their sum=15.
For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 [divisible by 3] only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.
Second step:
We have two set of numbers:
{1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets:
{1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.
{0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit [combination of 4] 4! --> 5!-4!=96
120+96=216
Answer: E.
_________________
Manager
Joined: 15 Sep 2009
Posts: 73
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
Only 2 sets are possible
case [1] 1,2,3,4,5
case [2] 0,1,2,4,5.
case [1] : there will 5! ways to form the number = 120
case [2] ; there will 4*4*3*2*1 = 96 ways
So total no.of ways = 120+96 = 216 ways
Senior Manager
Joined: 03 Sep 2006
Posts: 495
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
[#permalink]
By the property of divisibility by 3 i.e "a no: is divisible by 3, if the sum of the digits is divisible by 3"[e.g= 12-->1+2=3]
so from 0,1,2,3,4,5 the set of 5 digit no:s that can be formed which is divisible by 3 are 0,1,2,4,5[sum=12] & 1,2,3,4,5[sum=15]
from first set[0,1,2,4,5] no:s formed are 96 i.e first digit can be formed from any 4 no: except 0, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed.
from second set[1,2,3,4,5] no:s formed are 120 i.e first digit can be formed from any 5 digits, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed.
so total 120+96=216
Manager
Joined: 28 Aug 2010
Posts: 128
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 [divisible by 3] only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.
i understood the first part but did not get the second part 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. ..Could you please explain it in a little bit more detail. Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 86798
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
ajit257 wrote:
For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 [divisible by 3] only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.
i understood the first part but did not get the second part 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. ..Could you please explain it in a little bit more detail. Thanks
The sum of the given digits is already a multiple of 3
[15], in order the sum of 5 digits to be a multiple of 3 you must withdraw a digit which is itself a multiple of 3, otherwise [multiple of 3] - [non-multiple of 3] = [non-multiple of 3].
_________________
Manager
Joined: 28 Aug 2010
Posts: 128
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
so lets say we were asked a multiple of 5 so in that case we would have to withdraw the digit 5 ..is that correct ?
Math Expert
Joined: 02 Sep 2009
Posts: 86798
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
ajit257 wrote:
so lets say we were asked a multiple of 5 so in that case we would have to withdraw the digit 5 ..is that correct ?
5 or 0, as 0 is also a multiple of 5.
AGAIN: we have [sum of 6 digits]=[multiple of 3]. Question what digit should we withdraw so that the sum of the remaining 5 digits remain a multiple of 3? Answer: the digit which is itself a multiple of 3.
Below might help to understand this concept better.
If integers \[a\] and \[b\] are both multiples of some integer \[k>1\] [divisible by \[k\]], then their sum and difference will also be a multiple of \[k\] [divisible by
\[k\]]:
Example: \[a=6\] and \[b=9\], both divisible by 3 ---> \[a+b=15\] and \[a-b=-3\], again both divisible by 3.
If out of integers \[a\] and \[b\] one is a multiple of some integer \[k>1\] and another is not, then their sum and difference will NOT be a multiple of \[k\] [divisible by \[k\]]:
Example: \[a=6\], divisible by 3 and \[b=5\], not divisible by 3 ---> \[a+b=11\] and \[a-b=1\], neither is divisible by 3.
If integers \[a\] and
\[b\] both are NOT multiples of some integer \[k>1\] [divisible by \[k\]], then their sum and difference may or may not be a multiple of \[k\] [divisible by \[k\]]:
Example: \[a=5\] and \[b=4\], neither is divisible by 3 ---> \[a+b=9\], is divisible by 3 and \[a-b=1\], is not divisible by 3;
OR: \[a=6\] and \[b=3\], neither is divisible by 5 ---> \[a+b=9\] and \[a-b=3\], neither is divisible by 5;
OR: \[a=2\] and \[b=2\], neither is divisible by 4 ---> \[a+b=4\] and
\[a-b=0\], both are divisible by 4.
Hope it's clear.
_________________
Retired Moderator
Joined: 20 Dec 2010
Posts: 1252
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
[#permalink]
0,1,2,3,4,5
One digit will have to remain out for all 5 digit numbers;
if 0 is out; Leftover digits will be 1,2,3,4,5 =
Sum[1,2,3,4,5]=15.
5! = 120 numbers
if 1 is out; Leftover digits will be 0,2,3,4,5 = Sum[0,2,3,4,5]=14. Ignore[Not divisible by 3]
if 3 is out; Leftover digits will be 0,1,2,4,5 = Sum[0,1,2,4,5]=12.
4*4! = 4*24 = 96
if 4 is out; Leftover digits will be 0,1,2,3,5 = Sum[0,1,2,3,5]=11. Ignore
if 5 is out; Leftover digits will be 0,1,2,3,4 = Sum[0,1,2,3,4]=10. Ignore
Total count of numbers divisible by 3 = 120+96 = 216
Ans: "E"
Manager
Joined: 20 Aug 2011
Posts: 72
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
[#permalink]
A number is divisible by 3 if sum of its digits is a multiple of 3.
With the given set of digits, there are two possible combinations of 5 digits each-
A. [1,2,3,4,5] No. of possible 5 digit numbers: 5!= 120
B. [0,1,2,4,5] No. of possible 5 digit numbers: 4*4!=96 [the number can't start with a 0]
A+B= 120+96= 216
E
Originally posted by blink005 on 06 Jan 2012, 06:48.
Last edited by
blink005 on 06 Jan 2012, 08:27, edited 1 time in total.
Senior Manager
Joined: 13 Aug 2012
Posts: 356
Concentration: Marketing, Finance
GPA: 3.23
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?
A. 15
B. 96
C. 120
D. 181
E. 216
0 + 1 + 2 + 3 + 4 + 5 = 15
To form 5-digit number, we can remove a digit and the sum should still be divisible by 3.
15 - 1 = 14
15 - 2 = 13
15 - 3 = 12 BINGO!
15 - 4 = 11
15 - 5 = 10
Possible = {5,4,3,2,1} and {5,4,0,2,1}
There are 5! = 120 ways to arrange {5,4,3,2,1}
There are 5! - 5!/5 = 96 ways to arrange {5,4,0,2,1} since 0 cannot start the five number digit.
120 + 96 = 216
Answer: E
Manager
Joined: 12 Jan 2013
Posts: 55
Location: United States [NY]
GPA: 3.89
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
I did in 1 min 18 sec.
At first I wanted to choose a set of five digits, but started to worry about the complications with the leading
zero.
Then I thought that the last digits could always be chosen in only two ways so as to ensure divisibility by three - however, I quickly realized that I would not get all different digits.
Then I realized that once I get a number I can keep permuting the digits while still getting valid numbers.
In an attempt to avoid the leading zero I tried 12345 and noticed that it was divisible by 3. Thus, I've got 5!=120 answers and immediately eliminated two answers, A and B.
Then I addressed the case of a leading zero. Since I wanted to preserve divisibility by 3, I quickly saw that I could only use 0 instead of 3. Thus, the only other possible set was {0, 1, 2, 4, 5}. I tried adding another 5! and got 240, so the answer was slightly less than that.
After that I knew I had to subtract 4!=24 to account for all the possibilities with a leading zero, which left me with 240-24=216. This is how I do such problems...
_________________
Sergey Orshanskiy,
Ph.D.
I tutor in NYC: //www.wyzant.com/Tutors/NY/New-York/7948121/#ref=1RKFOZ
Intern
Joined: 09 Jul 2013
Posts: 20
Location: United States [WA]
GPA: 3.65
WE:Military Officer [Military & Defense]
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
E.
I'm offering up a way to apply the "slot method" to this problem below.
First, as everyone else has identified above, you need to find the cases where the 6 numbers [0, 1, 2, 3, 4, 5] create a 5-digit number divisible by 3.
Shortcut review: a number is divisible by 3 if the sum of the digits in the number is divisible by 3.
So, 12345 would be divisible by 3 [1+2+3+4+5 = 15, which is divisible by 3].
Analyzing the given numbers, we can conclude that only the following two groups of numbers work: 1, 2, 3, 4, 5 [in any order, they would create a five digit number divisible by 3 - confirmed by the shortcut above], and 0, 1,2, 4, 5.
Now we need to count the possible arrangements in both cases, and then add them together.
To use the slot method with case 1 [1,2,3,4,5]:
_ _ _ _ _ [five digit number, 5 slots]. Fill in the slots with the number of "choices" left over from your pool of numbers. Starting from the left, I have 5 choices I can put in slot #1 [5 numbers from the group 1,2,3,4,5 - pretend I put in number 1, that leaves 4 numbers]
5 _ _ _ _
Fill in the next slot with the
number of choices left over [4 choices left...numbers 2 through 5]
5 4 _ _ _
Continue filling out the slots until you arrive at:
5 4 3 2 1
Multiply the choices together: 5x4x3x2x1 [which also happens to be 5!] = 120 different arrangements for the first case.
Now consider case 2 [0,1,2,4,5]:
_ _ _ _ _ [five digit number, 5 slots]. Here's the tricky part. I can't put 0 as the first digit in the number...that would make it a 4-digit number! So I only have 4 choices to pick from
for my first slot!
4 _ _ _ _
Fill in the next slot with the remaining choices [if I put in 1 in the first slot, I have 0,2,4,5 left over...so 4 more choices to go].
4 4 _ _ _
Continue to fill out the slots with the remaining choices:
4 4 3 2 1
Multiply the choices together: 4 x 4! = 96 different arrangements for the second case.
Now the final step is to add all the possible arrangements together from case 1 and case 2:
120 + 96 = 216. And this is our answer.
Hope this alternate "slot" method helps! This is how I try to work these combinatoric problems instead of using formulas... in this case it worked out nicely. Here, order didn't matter [we are only looking for total possible arrangements] in the digits, so we didn't need to divide by the factorial number of slots.
Senior Manager
Joined: 07 Apr 2012
Posts: 277
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
Bunuel wrote:
TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?
A. 15
B. 96
C. 120
D. 181
E. 216
First step:
We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.
We have six digits: 0,1,2,3,4,5. Their sum=15.
For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 [divisible by 3] only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.
Second step:
We have two set of numbers:
{1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets:
{1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.
{0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit [combination of 4] 4! --> 5!-4!=96
120+96=216
Answer: E.
I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers divisible by 3.
What is wrong with this logic?
Math Expert
Joined: 02 Sep 2009
Posts: 86798
Re: How
many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
ronr34 wrote:
Bunuel wrote:
TheRob wrote:
How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?
A. 15
B. 96
C. 120
D. 181
E. 216
First step:
We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.
We have six digits: 0,1,2,3,4,5. Their sum=15.
For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 [divisible by 3] only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.
Second step:
We have two set of numbers:
{1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets:
{1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.
{0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit [combination of 4] 4! --> 5!-4!=96
120+96=216
Answer: E.
I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers divisible by 3.
What is wrong with this logic?
Because the numbers divisible by 3 are not 1/3rd of all possible numbers.
{0, 1, 2, 3, 4} --> 96 5-digit
numbers possible with this set.
{0, 1, 2, 3, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set.
Total = 5*5*4*3*2 = 600 but the numbers which are divisible by 3 come
from third and sixth sets: 96 + 120 = 216.
_________________
GMAT Expert
Joined: 16 Oct 2010
Posts: 13164
Location: Pune, India
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
ronr34 wrote:
I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers
divisible by 3.
What is wrong with this logic?
We cannot do this because we have the asymmetric 0 as one of the digits. The number of 5 digit numbers that can be formed with 0, 1, 2, 3 and 4 is different from the number of 5 digit numbers that can be formed with 1, 2, 3, 4 and 5 [because 0 cannot be the first digit].
Had the digits been 1, 2, 3, 4, 5 and 6, then your method would have been correct.
If 0 is included:
{0, 1, 2, 3, 4} --> 96 5-digit numbers
possible with this set.
{0, 1, 2, 3, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 4, 5} --> 96 5-digit numbers possible with this set. - All these numbers are divisible by 3
{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
The number of 5 digit numbers in these sets is
not the same - Sets with 0 have fewer numbers
If 0 is not included:
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 4, 6} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 5, 6} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
{1, 2, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{1, 3, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{2, 3, 4, 5, 6} --> 120 5-digit
numbers possible with this set. - All these numbers are divisible by 3
Here exactly 1/3rd of the numbers will be divisible by 3.
_________________
Karishma
Owner of Angles and Arguments at //anglesandarguments.com/
NOW PUBLISHED - DATA SUFFICIENCY MODULE
For Individual GMAT Study Modules, check Study Modules >
For Private Tutoring, check
Private Tutoring >
CrackVerbal Representative
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4989
Location: India
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
[#permalink]
We need to form five digit numbers with distinct digits which are divisible by 3. A number is divisible by 3 when the sum of its digits is divisible by 3.
When we observe the digits, we see that the sum of the given digits is 15. Typical of a GMAT kind of question – data given is very precise and rarely vague.
So, there are only two cases that can be considered to fit the constraints given
Case 1: A 5 digit number with the digits {1,2,3,4,5}. Since 0 is not a part of this set and there are 5 different digits, we can form a total of 5P5 = 5! = 120 numbers. All of these will be divisible by 3.
At this stage, we can eliminate answer options A, B and C.
Case 2: A 5 digit number with the digits {0,1, 2, 4, 5}. Since 0 is a part of this set, we need to use Counting methods to find out the number of 5-digit numbers.
The ten thousands place can be filled in 4 ways, since 0 cannot come here; the thousands place can be filled in 4 ways, the hundreds in 3 ways, the tens in 2 ways and the units place in 1 way.
Therefore, number of 5-digit numbers = 4 * 4 * 3 * 2 * 1 = 96.
Total number of 5-digit numbers with distinct digits, divisible by 3 = Case 1 + Case 2 = 120 + 96. Answer option D can be eliminated.
The correct answer option is E.
Hope that helps!
Aravind B T
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 24390
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable [esp those replies with Kudos].
Want to see all other topics I dig out? Follow me [click follow button on profile]. You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: How many five digit numbers can be formed using digits 0, 1, 2, 3, 4 [#permalink]
01 Jul 2022, 03:04
Moderators:
Senior Moderator - Masters Forum
3084 posts