Latest Elitmus Aptitude Question SOLUTION: what is the sum of all 5 digits numbers which can be formed with the digits 0,1,2,3,4 without repetition

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(#M40155271) ELITMUS Numerical Ability question number Keep an EYE

what is the sum of all 5 digits numbers which can be formed with the digits 0,1,2,3,4 without repetition

Asked In Elitmus ROHIT KUMAR DUBEY (8 years ago)

Solved Bhavya Dhawal Read Solution (10)

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Read Solution (10) :

Bhavya Dhawal  (8 years ago) Best Solution

There is a simplest formula for solving these type of questions when we are using numbers along with 0 for finding sum.
i,e........(sum of numbers)*{(n-1)!*(111...........n times)-(n-2)!*(111..........(n-1)times) }
so,here by putting the values as,

(0+1+2+3+4){(5-1)!*(11111)-(5-2)!*1111}

= 10{(4!*11111)-(3!*1111)}

= 10{266664-6666}

= 2599980 ans

Is this solution Helpfull? Yes (66) | No (5)  

paras  (8 years ago)

If u observe each digit will be present at each position in 4! time.
i.e., No of times 0 is in first(ten thousand) place is (5-1)! = 4!= 24 ways.
In the same way ,
No of times 1 is in first place is (5-1)! = 4!= 24 ways.
No of times 2 is in first place is (5-1)! = 4!= 24 ways.
No of times 3 is in first place is (5-1)! = 4!= 24 ways.
No of times 4 is in first place is (5-1)! = 4!= 24 ways.
No of times 0 is in second( thousand) place is (5-1)! = 4!= 24 ways.
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.
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No of times 4 in fifth(One's) place is (5-1)! = 4!= 24 ways.

So the sum of all numbers will be
= (5-1)! * (10000+1000+100+10+1)*(0+1+2+3+4)
=4! * (11111)*10
= 24* 11111*10
= 2666640

Logic can be (n-1)!*(111...1 ntimes)*(sum of digits)


As numbers formed by digits 1,2,3,4 and 0 as first digits are not 5 digit numbers, so subtracting 4 digit numbers formed by 1,2,3,4
i.e., 2666640 - 3!*(1111)*(1+2+3+4)
= 2666640 - 6666*10
= 2666640 -66660
= 2599980

Is this solution Helpfull? Yes (18) | No (1)  

ROHIT KUMAR DUBEY  (8 years ago)

no bhavya if 0 is included excluded then formulae goes as
.(sum of numbers)*{(n-1)!*(111...........n times) only ,u r formulae is for the Q like if the digit is given 1,2,3,4,5 and we have to make find sum of number form of 2 digit from these numbers. then u r formulae is applicable. see once. visit this link:-https://www.youtube.com/watch?v=leBWTcqACkg

Is this solution Helpfull? Yes (8) | No (3)  

richa  (8 years ago)

total no. of 5 digit numbers formed by 0,1,2,3,4 are 4*4!=96 .because at first place 0 cant be placed .
so,numbers formed when 1 at first place=4!
numbers formed when 2 at first place=4!
numbers formed when 3 at first place-4!
numbers formed when 4 at first place=4!
total =4*4!=96.
now sum of all the numbers ,
at 10000th place each number will come 24 times except 0
at 1000th place each number will come 24 times
and so on ...........
at 10000th place =24*(1+2+3+4)=24*10 =2,40,00,00 (at 10,000th place)
at 1000th place=24*(0+1+2+3+4+5)=24 *10=2,40,000
at 100th place=24*(0+1+2+3+4+5)=24*10=2,40,00
at 10th place=24*(0+1+2+3+4+5)=24*10=2400
at unit place =24*(0+1+2+3+4+5)=240
so sum of the numbers =2400000+240000+24000+2400+240=2666640

Is this solution Helpfull? Yes (1) | No (5)  

Bhavya Dhawal  (8 years ago)

If 0 is not included in the question then,

sum of the given numbers*P(n-1,r-1)(1111.........n times)

where n= nos. of the digits from which we have to find sum,

r= sum of the given digits.

Is this solution Helpfull? Yes (1) | No (6)  

  (4 years ago)

If 0 is present in given digits then formula is :
(sum of numbers)*{(n-1)!*(111...........n times)-(n-2)!*(111..........(n-1)times) }..
and if 0 is not present in given digits then formula is:
(sum of numbers)*{(n-1)!*(111...........n times)..
In this case,since 0 is present in the given digits so we will be using (sum of numbers)*{(n-1)!*(111...........n times)-(n-2)!*(111..........(n-1)times) }..
and solution is:
(0+1+2+3+4)*(5-1)!*(11111)-(1+2+3+4)*(5-2)!*(1111)

=10*4!*11111-10*3!*1111

=2666640-66660

=2599980

Is this solution Helpfull? Yes (1) | No   

Mohd Jahid  (4 years ago)

There are total 96 numbers. Sum of single number obviously be 0+1+2+3+4=10.
Required Sum is 96*10=960.

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ROHIT KUMAR DUBEY  (8 years ago)

bhavya if 0 is exluded then what will be the method to solve this.

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keerthi  (7 years ago)

(0+1+2+3+4)*(4!)(11111)

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Dwarkesh  (7 years ago)

Consider the string 123456789. Observe that each of the 99 digits in the string can safely be either included or not included, since failing to include a digit will not violate the increasing order of the digits. Hence, there are 2^9 possible strings that contain (0 or 1 or 2 or ... or 9) digits. However, we likely want to omit the case where there are 00 digits, which leaves us with a final answer of:
2^9−1=511
:)

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