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(#M40157000) ELITMUS Numerical Ability question P and C Keep an EYE

What is the sum of all 3 digits number that can be formed using digits 0,1,2,3,4,5 with no repitition,

Asked In Elitmus Sagar Wadjikar (8 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

Jaya Ganthan  (8 years ago)

We know that zero can't be in hundreds place. But let's assume that our number could start with zero.

The formula to find sum of all numbers in a permutation is

111 * no of ways numbers can be formed for a number at given position * sum of all given digits

No of 1 s depends on number of digits

So,the answer us

111 * 20 *(0+1+2+3+4+5) = 33300

We got 20 as follows..if we have 0 in units place we can form a number in 4*5 ways...this is for all numbers. So we have substituted 20 in formula..hope u understood

Now, this is not the final answer because we have included 0 in hundreds place...so we have to remove the sum of all numbers that starts with 0.

This is nothing but the sum of all 2 digits numbers formed by 1 2 3 4 5.. because 0 at first place makes it a 2 digit number.

So the sum for this is 11 * 4 * (1+2+3+4+5).
=660

Hope u understood why we use 4..each number can be formed in 4*1 ways

So, the final answer is 33300-660 = 32640

Is this solution Helpfull? Yes (45) | No (6)  

Shubham Brat  (8 years ago)

total no of 3 digit numbers possible are 5*5*4=100
sum of all 3digit no= 111*100(0+1+2+3+4+5)=11100*15=166500

Is this solution Helpfull? Yes (11) | No (3)  

Rajeev kumar  (7 years ago)

number of digit given is n(including 0)
number formed by r digit
formula
sum of all r digit is
n-1 p r-1(1111.....r times)(sum of digits)-n-2 p r-2(1111.......r-1 times)(sum of digits)
here,
n=6(0,1,2,3,4,5)
r=3
sum is
5 p 3(111)(0+1+2+3+4+5) - 4 p 1(11)-(0+1+2+3+4+5)
20(111)*15 -4(11)*15
300(111)-60(11)
32640

Is this solution Helpfull? Yes (4) | No   

KUMAR HARSH  (8 years ago)

There are 5*5*4=100 way
If you pick one at random, the expected value of the first digit is (1+2+3+4+5)/5=3
and the expected value of each other digit must be 12/5
because the sum of all five digits is 0+1+2+3+4+5=3+5*12/8
0+1+2+3+4=52+4⋅158. By linearity of expectation, the sum of all 100 numbers is 100(3*100+12/5*11)=
32640 is ans

Is this solution Helpfull? Yes (2) | No (2)  

rajat  (8 years ago)

we know that number of way to organise a number in a way..
for above question..
suppose we fix the 5 at the end of number_ _ 5.then we have to find the number of way to find the 2 digit number including 0 either they appear in tens,hundread,unit place..for 2 number number of way
0,1,2,3,4,5=number of way =5*4=20.
for total sum including 0 each number 20 time in unit or hundread place so...
total sum(0+1+2+3+4+5)*(20+200+2000).
for unit =20 for tenth =20*10=200
for hundread 200*10=2000.
total=(0+1+2+3+4+5)*(20+200+2000_)=33300.
and subtract number statt with zero..
fix the 5 at end and does not consider 0from 0,1,2,3,4,5.
_ _5 number of way4*3=12 way.
(1+2+3+4+5)(12+120)=1980.
sum=33300-1980=31320.

Is this solution Helpfull? Yes (1) | No (4)  

prince pandey  (8 years ago)

Total numbers that can be formed= 5*5*4=100
How many time 0 will appear at unit place?
5*4=20 (Since 0 is at unit place)
How many times 1,2,3,4,5 will comes at unit place?
Lets take 1 at unit place. So, 0 can not be at leading position.
4*4=16
Similarly, 2,3,4,5 will appear 16 times at units position.
So, Sum of Unit place digit:
16*(1+2+3+4+5)=16*15=240
Same number of times all numbers will appear at tens place and hundreds place. So,
100*240+10*240+1*240=24000+2400+240=26640

Is this solution Helpfull? Yes (1) | No (4)  

SAURAV KUMAR  (8 years ago)

3 digit number without repetition formed as
102, 103, 104, 105, 203, 204, 205, 304, 305, 405
now we have to find sum of all above no...........
102-> 2!*111*3
103->2!*111*4
.
.
.
.405->2!*111*9
now sum of all number will be=2!*111*(3+4+5+6+5+6+7+7+8+9)
= 2!*111*60
=13410
this will be the answer

Is this solution Helpfull? Yes (1) | No (2)  

pooja kumari  (8 years ago)

@jaya why multiply with 111?

Is this solution Helpfull? Yes | No (1)  

madhav  (8 years ago)

@Anjali if repeatation is not allowed then answer will be 8, if allowed answer will be 12.
no is 345,375,435,465,645,675,735,765

Is this solution Helpfull? Yes | No   

lohit  (8 years ago)

If you fix 5 as the last digit, you see that there are 4⋅4⋅ways to complete the number. Thus, 5 appears 16 times as the last digit. By the same logic, if we enumerate all possible numbers using these 6 digits, each number appears 16 times in each of the 3 positions. That is, the digit 5 contributes addition of total numbers In total, we have
(0+1+2+3+4+5)(16+160+1600)= 26640

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akshay  (7 years ago)

@jaya ganthan how 4 has come in two digit no

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Ayush  (7 years ago)

total no of 3 digit numbers possible are 5*5*4=100 therefore n=100
First term can be 102 =a
Last Term can be 543 =l
So using Sum=n/2[a+l]=100/2[102+543]
ans=32250

Is this solution Helpfull? Yes | No (1)  

sooraj singh  (7 years ago)

best solution @rajeev kumar.. simple and effective approach .

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  (6 years ago)

sum of unit digits of all the no=(total no of digits that can be placed on hundred place)*(total no of digits that can be placed on tenth place)*(0+1+2+3+4+5)=4*4*15

sum of tenth digits of all the no=(total no of digits that can be placed on hundred place)*(0+1+2+3+4+5)*(total no of digits that can be placed on unit place)=4*15*4

sum of hundred digits of all the no=(1+2+3+4+5)(total no of digits that can be placed on hundred place)**(total no of digits that can be placed on unit place)=15*5*4

note-[ 0 cant be placed on hundred place, hence0 not added in the last ] also [for this it is now 5*4 not 4*4]
now you can add up and take carries .
ans is 32640

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