Mẹo về How many numbers of 3 digits can be formed with the digits 1 2 3 4 and 5 without any repetition of digits? Chi Tiết
Dương Văn Hà đang tìm kiếm từ khóa How many numbers of 3 digits can be formed with the digits 1 2 3 4 and 5 without any repetition of digits? được Cập Nhật vào lúc : 2022-09-19 10:34:09 . Với phương châm chia sẻ Bí quyết về trong nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi Read nội dung bài viết vẫn ko hiểu thì hoàn toàn có thể lại Comment ở cuối bài để Admin lý giải và hướng dẫn lại nha.For a number to be divisible by $6$, it must be divisible by both $2$ and $3$. If it is divisible by $2$, it must be even, so the units digit must be $2$ or $4$. If it is divisible by $3$, the sum of its digits must be divisible by $3$.
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The only one-digit positive integer that is divisible by $6$ is $6$ itself, so the number must have least two digits.
Two-digit numbers: If the units digit is $2$, the tens digit must have remainder $1$ when divided by $3$. Hence, the tens digit must be $1$ or $4$.
If the units digit is $4$, the tens digit have remainder $2$ when divided by $3$. Hence, the tens digit must be $2$ or $5$.
Therefore, there are four two-digit numbers divisible by $6$ that can be formed using the digits $1, 2, 3, 4, 5$ without repetition. They are $12$, $24$, $42$, $54$.
Three-digit numbers: If the units digit is $2$, the sum of the hundreds digit and tens digit must have remainder $1$ when divided by $3$. Since the sum of the hundreds digit and tens digit must be least $1 + 3 = 4$ and most $5 + 4 = 9$, the only possibilities are that the sum of the hundreds digit and tens digit is $4$ or $7$. Since digits cannot be repeated, the only way to obtain $4$ is to use the digits $1$ and $3$ in either order, and the only way to obtain $7$ is to use the digits $3$ and $4$ in either order. Hence, there are four three-digit numbers divisible by $6$ that can be formed with the digits $1, 2, 3, 4, 5$ that have units digit $2$. They are $132$, $312$, $342$, and $432$.
If the units digit is $4$, then the sum of the hundreds digit and tens digit must have remainder $2$ when divided by $3$. Since the sum of the hundreds digit and tens digit must be least $1 + 2 = 3$ and most $3 + 5 = 8$, the sum of the hundreds digit and tens digit must be $5$ or $8$. Since digits cannot be repeated, the only way to obtain $5$ is to use the digits $2$ and $3$ in either order, and the only way to obtain $8$ is to use the digits $3$ and $5$ in either order. Hence, there are also four three-digit numbers divisible by $6$ that can be formed with the digits $1, 2, 3, 4, 5$ that have units digit $4$. They are $234$, $324$, $354$, $534$.
Therefore, there are a total of eight three-digit numbers divisible by $6$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition.
Four-digit numbers: If the units digit is $2$, then the sum of the thousands digit, hundreds digit, and tens digit must have remainder $1$ when divided by $3$. Since the sum of the thousands digit, hundreds digit, and tens digit must be least $1 + 3 + 4 = 8$ and most $3 + 4 + 5 = 12$, the sum of the thousands digit, hundreds digit, and tens digit must be $10$. Since digits cannot be repeated, the only way to obtain a sum of $10$ is to use the digits $1$, $4$, and $5$ in some order. There are $3! = 6$ such orders. Hence, there are six four-digit numbers divisible by $6$ with units digit $2$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition. They are $1452$, $1542$, $4152$, $4512$, $5142$, and $5412$.
If the units digit is $4$, the remainder of the sum of the thousands digit, hundreds digit, and tens digit must be $2$ when divided by $3$. Since the sum of the thousands digit, hundreds digit, and tens digit must be least $1 + 2 + 3 = 6$ and most $2 + 3 + 5 = 10$, the sum of the thousands digit, hundreds digit, and tens digit must be $8$. Since digits cannot be repeated, the only way to obtain a sum of $8$ is to use the digits $1$, $2$, and $5$ in some order. Since there are $3! = 6$ such orders, there are also six four-digit numbers that can be formed from the digits $1, 2, 3, 4, 5$ without repetition. They are $1254$, $1524$, $2154$, $2514$, $5124$, and $5214$.
Hence, there are a total of $12$ four-digit numbers divisible by $6$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition.
Five-digit numbers: The sum of the five digits $1, 2, 3, 4, 5$ is $15$, which is divisible by $3$. Hence, any five digit number formed from these digits without repetition that has units digit $2$ or $4$ is divisible by $6$. There are two ways of filling the units digit and $4!$ ways of filling the remaining digits. Hence, there are $2 cdot 4! = 48$ five-digit numbers that can be formed with the digits $1, 2, 3, 4, 5$ without repetition.
In total, there are $4 + 8 + 12 + 48 = 72$ numbers divisible by $6$ that can be formed from the digits $1, 2, 3, 4, 5$ without repetition.
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How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition) [#permalink]
Updated on: 20 Jan 2022, 05:37
00:00
Question Stats:
61%
(02:07) correct 39% (02:19) wrong based on 109 sessions Hide Show timer Statistics
How many five digit numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit the unit’s place must be greater than that in the ten’s place?
(a) (54)
(b) (60)
(c) (17)
(d) (2 × 4!)
(e) (120)
Originally posted by sharathnair14 on 10 Jan 2022, 09:38.
Last edited by
sharathnair14 on 20 Jan 2022, 05:37, edited 1 time in total.
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How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition) [#permalink]
Updated on: 15 Jul 2022, 09:45
Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digitsi) = 5*4*3*2*! = 5! =
120.
Now, in half them unit's digit will be bigger than the ten's digit and in half of them it will be smaller.
Example: Let's say we have three digits 1,2,3. Total number of numbers without repeating digits = 3*2*1=6
Numbers with Unit's digit greater than the ten's digit
123, 213, 312
Numbers with Ten's digit greater than the unit's digit
321, 132, 231
So total Number of cases = 120/2 = 60
So, Answer will be B
Hope it helps!
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Originally posted by BrushMyQuant on 11 Jan 2022, 08:59.
Last edited by BrushMyQuant on 15 Jul 2022, 09:45, edited 1 time in total.
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Re: How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition) [#permalink]
19 Jan 2022, 01:24
sharathnair14 wrote:
How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit the unit’s place must be greater than that in the ten’s place?
(a) (54)
(b) (60)
(c) (17)
(d) (2 × 4!)
(e) (120)
unit's place>ten's place
So , possible unit digit = 2.3.4.5
when 2 is in unit's digit 1 must be in ten's and (3,4,5) forms the other numbers.
total possible number =3!=6
similarly when 3 is in unit's digit 1 or 2 can be in ten's digit and 3 other digits form the number.
so total possible number =3!*2=12
again when 4 ................. total possible number =3!*3=18
and when 5 .................. total possible number =3!*4=24
sum of total possibilities =6+12+18+24=60
Answer: B
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Re: How many numbers can be formed from 1, 2, 3, 4, 5
(without repetition) [#permalink]
19 Jan 2022, 03:42
Does it mean a five digit number? A number can be 2 digit , 3 digit till 5 digit for this combination
Posted from my mobile device
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Posts: 86786
Re: How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition) [#permalink]
19 Jan 2022, 03:46
ManjariMishra wrote:
Does it mean a five digit number? A number can be 2 digit , 3 digit till 5 digit for this combination
Posted from my mobile device
You are right. The question should mention that we are looking for 5-digit numbers only.
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Re: How many numbers can be
formed from 1, 2, 3, 4, 5 (without repetition) [#permalink]
19 Jan 2022, 04:05
Condi-1:Digit unit place> digit tens place.
Condi-2: Without repetition
(1,2,3,4,5)
possible combinations for tens place and unit place, 5C2= 10. Here we will not multiply by 2! because we want ascending order. For example, (2,1) and (1, 2) are two pair but we need only (2,1) which is satisfying condition-1
For remaining places, arrangement of remaining digits is 3*2*1= 6.
So total ways of arrangement= 6*10= 60.
B is answer.
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Re: How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition)
[#permalink]
02 Feb 2022, 22:12
sharathnair14 wrote:
How many five digit numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit the unit’s place must be greater than that in the ten’s place?
(a) (54)
(b) (60)
(c) (17)
(d) (2 × 4!)
(e) (120)
No of 5 digit numbers with 1, 2, 3, 4, 5 digits = 5! = 120
By symmetry, in half of them, the units digit will be greater that tens digit and in the other half, the tens digit will be greater than units digit.
So 120/2 = 60
Answer (B)
Note the symmetry - If 1 is in units digit, all such numbers will not be included. If
5 is in the units digit, all such numbers will be included. If 2 is in units digit, only numbers with 1 is tens digit will be included. If 4 is in units digit, only number with 5 in tens digit will not be included. When 3 is in units digit, half the numbers will be acceptable and half will not be.
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Re: How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition) [#permalink]
02 Feb 2022, 22:12
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How many numbers of 3 digits can be formed with the digits 1 2 3 4 5 repetition of digits are not allowed )?
Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? Solution: Answer: 108. Let 3-digit number be XYZ.How many 3
so 60(ans.)How many 3
Solution : (i) When repetition of digits is allowed:No. of ways of choosing firsy digits = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
Therefore, total possible numbers `= 5 xx 5 xx 5 = 125`
(ii) When repetition of digits is not allowed:
No.
How many ways can a 3
So the number of 3-number combinations are- (1,2,3),(1,2,4),(1,3,4),(2,3,4). Each can be arranged in 6 ways, so we get 24 ways totally. So why is my answer different here? Tải thêm tài liệu liên quan đến nội dung bài viết How many numbers of 3 digits can be formed with the digits 1 2 3 4 and 5 without any repetition of digits?