Let´s assume that we don´t have to use all digits 1,2 and 3 to form the number. We define $X,Y,Z\in \{1,2,3 \}$
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- How many 5-digit prime numbers can be formed using the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
- Answer (Detailed Solution Below)
- The sum of all five-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 when repetition of digits is not allowed, is
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- How many numbers of 5 digits can be formed using the digits 1 2 3 4 5 6 such that digits are repeated?
- How many 5 digit numbers can be formed using the digits 1 2 3 4 and 5 Repetition not allowed?
- How many 5 digit numbers can be formed from the digits 1 2 3 4 5 and 6 such that the numbers are divisible by 4 and their digits do not repeat?
- How many 5 digit numbers can be made from the digits 1 2 3 4 so that all the digits are taken in each number?
- How many 5 digit numbers can be formed using digits 1,2 3 4 and 5 if repetition of digits is not allowed?
- How many 5 digit numbers can be formed from the digits 1,2 3 4 5 and 6 such that the numbers are divisible by 4 and their digits do not repeat?
- How many 5 digit numbers can be formed using the digits 0 1,2 3 4 and 5 which are divisible by 5 without repetition of the digits?
- How many numbers of 5 digits can be formed with the digits 0 1,2 3 4 if the digits can repeat themselves?
Nội dung chính
- How many 5-digit prime numbers can be formed using the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
- Answer (Detailed Solution Below)
- The sum of all five-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 when repetition of digits is not allowed, is
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- How many numbers of 5 digits can be formed using the digits 1 2 3 4 5 6 such that digits are repeated?
- How many 5 digit numbers can be formed using the digits 1 2 3 4 and 5 Repetition not allowed?
- How many 5 digit numbers can be formed from the digits 1 2 3 4 5 and 6 such that the numbers are divisible by 4 and their digits do not repeat?
- How many 5 digit numbers can be made from the digits 1 2 3 4 so that all the digits are taken in each number?
Case 1: Two different groups of digits, where each group consists of one kind of number.: $XXXYY$
These sequence can be arranged in $\frac{5!}{3!\cdot 2!}=\frac{120}{12}=10$ ways.
$X$ and $Y$ can have the following combinations: $(1,2);(2;1);(1,3);(3,1);(2,3);(3,2)$
Thus for case 1 we have $6\cdot 10=60$ ways.
Case 2: Three different groups of digits where each group consists of one kind number and one group has 3 digits: $XXXYZ$. This sequence can be arranged in $\frac{5!}{3!\cdot 1!\cdot 1!}=\frac{120}{6}=20$ ways.
$X,Y,Z$ can have $3$ combinations and for case 2 there exists $60$ ways.
Finally we can say that $120$ five digit numbers can be formed using digits 1,2,3 with exactly one digit repeating 3 times.
How many 5-digit prime numbers can be formed using the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
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- 4
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Answer (Detailed Solution Below)
Option 4 : 0
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Concept:
Prime number: Prime number are those which are divisible by itself and 1.
Calculation:
Not a single five-digit prime number can be formed using the digits 1, 2, 3, 4, 5(without repetition).
This is because if one adds the digits, the result obtained will be = 1 + 2 + 3 + 4 + 5 = 15 which is divisible by 3.
Hence, any number obtained as a permutation of these 5 digits will be at least divisible by 3 and cannot be a prime number.
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The sum of all five-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 when repetition of digits is not allowed, is
(1) 366000
(2) 660000
(3) 360000
(4) 3999960
Answer: (4) 3999960
Solution:
Sum of all 5- digit numbers by using 1,2,3,4,5 without repetition=
(Sum of all digits) (n-1)! (10n-1/10-1)
= (1+2+3+4+5)4!(105-1/10-1)
= 15 x 4 x 3 x 2 x (105-1)/(10-1)
= 15 x 12 x 2 (100000-1)/9
= 40 x 99999
= 399960
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How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition) [#permalink]
Updated on: 20 Jan 2020, 05:3700:00
Question Stats:
61% (02:07) correct 39% (02:19) wrong based on 109 sessionsHide Show timer Statistics
How many five digit numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at 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 2020, 09:38.
Last edited by sharathnair14 on 20 Jan 2020, 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 2020, 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 2020, 08:59.
Last edited by BrushMyQuant on 15 Jul 2020, 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 2020, 01:24sharathnair14 wrote:
How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at 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 2020, 03:42Does 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: 86794
Re: How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition) [#permalink]
19 Jan 2020, 03:46ManjariMishra 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 2020, 04:05 Condi-1:Digit at unit place> digit at 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 2021, 22:12sharathnair14 wrote:
How many five digit numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at 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 2021, 22:12
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How many numbers of 5 digits can be formed using the digits 1 2 3 4 5 6 such that digits are repeated?
The answer is 120 five-digit integers (5!=
How many 5 digit numbers can be formed using the digits 1 2 3 4 and 5 Repetition not allowed?
dat can b formed using 1,2,3,4,5 when repetitions is not allowed? Now the total no of possible ways in which these 5 no can be written without repetition are 5!= 120.
How many 5 digit numbers can be formed from the digits 1 2 3 4 5 and 6 such that the numbers are divisible by 4 and their digits do not repeat?
Therefore, there are a total of 192 numbers which can be formed using the digits 1,2,3,4,5,6 without repetition such that the number is divisible by 4.
How many 5 digit numbers can be made from the digits 1 2 3 4 so that all the digits are taken in each number?
For each such 4-digit number we can form we have 5 choices for the first digit (which cannot be 0), and with repetition allowed we have 6 choices for each of the last 3 digits. Therefore, we can form 5*6*6*6 = 1080 such 4-digit numbers.
How many 5 digit numbers can be formed using digits 1,2 3 4 and 5 if repetition of digits is not allowed?
hence the possibilities are 4×9×9×9=2916 numbers are possible. You listed 7 digits, each number with 5 positions, and that repetitions are allowed. So, this is simply 7⁵, which is 16,807.
How many 5 digit numbers can be formed from the digits 1,2 3 4 5 and 6 such that the numbers are divisible by 4 and their digits do not repeat?
Therefore, there are a total of 192 numbers which can be formed using the digits 1,2,3,4,5,6 without repetition such that the number is divisible by 4.
How many 5 digit numbers can be formed using the digits 0 1,2 3 4 and 5 which are divisible by 5 without repetition of the digits?
Answer : `=2xx5xx6xx6xx6=2160` ways. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
How many numbers of 5 digits can be formed with the digits 0 1,2 3 4 if the digits can repeat themselves?
Since, repetition is not allowed, for the next significant place, 4 digits are available (since 0 can now be used). Sum of the digits = 0 + 1 + 2 + 3 + 4 = 10 which is not divisible by 3. ∴ None of the 5-digit numbers formed using the digits 0, 1, 2, 3, and 4 will not be divisible by 3.