How many 5 digit numbers can be formed if each one uses all the digits 0, 1,2, 3, 4 without repetition
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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
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Manager Joined: 29 May 2008 Posts: 91 How many five digit numbers can be formed using digits 0, 1, 2, 3, 4
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Hide Show timer StatisticsHow 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 Originally posted by TheRob on 22 Oct 2009, 13:20. 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 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} --> 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 (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
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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
\(k\)): 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 \(k\)): 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 \(k\)): 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
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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. 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. if 4 is out; Leftover digits will be 0,1,2,3,5 = Sum(0,1,2,3,5)=11. 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
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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 A+B= 120+96= 216 E Originally posted by blink005 on 06 Jan 2012, 06:48. 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 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 Possible = {5,4,3,2,1} and {5,4,0,2,1} There are 5! = 120 ways to arrange {5,4,3,2,1} 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. 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. 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): Now consider case 2 (0,1,2,4,5): Now the final step is to add all the possible arrangements together from case 1 and case 2: 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 First step: 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} --> 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: 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 First step: 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} --> 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: 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. 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: 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: If 0 is not included: Karishma For Individual GMAT Study Modules, check Study Modules > 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
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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. 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! 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 How many 5∴ the total number of arrangements = 120 + 96 + 96 = 312.
How many combinations are there with 5 numbers without repetition?How many combinations with 5 numbers without repetition are possible? There's one (1) possible combination without repetitions C(n,r) and 126 combinations with repetitions C'(n,r) of arranging a group of five numbers (i.e., the 1-5 number list).
How many 5But the answer is 90.
How many numbers of 5 digits can be formed using all digits of each digit occurs once?Therefore, the total number of combinations possible are 10 × 10 × 10 × 10 × 10 = 1,00,000. Out of these 1,00,000 ways of forming a 5-digit number, we need to take out all the possibilities which have zero in the first position on the left (or the ten thousand place).
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