**Question:**

How many 4 digit numbers are possible using the digits 0,1,2,3,4.. Find the sum of all those numbers...

No. Of ways to pick 4 out of 5 = 5C4 = 5

Nomber of ways 4 digits ca be rearranged = 4! = 4×3×2×1 =24

Total nomber of ways that 4 digited number from 5 digits are

5x24 = 120

But here 0 is given in the digit

So total So number of ways 0 at 1000's place

4 ways include 0 to select 4 digits from 5 digits

No of ways 0 can he at 1000's place for selected 4 digits which include 0 = 3! = 3×2×1 = 6

Total ways 0 comes in 1000's place = 4×6 = 24

So total number of ways 120-24 = __96__

Question doesnt clear for repetetion so we can repet digit for eg. 1111, 1011,1234,1223 and others it means first choice 4, second choice 5, third 5, and fourth also 5

So 4×5×5×5=500

We can't write down the zero at first.. So, for first digit we have only 4 option remain..but for second digit again we have 4 option coz now we can put zero ..for third digit we have 3 option and 2 option for last digit(digit of unity)..

So we can write it by..4×4×3×2= 96

Anyways thanks for pointing mah mistake.....

Ohhh yup sorry just forget to do this

@himanshu

Ur approach is correct.. but when u have 4 4 3 2... we dont add them.. we multiply them to get to total possible numbers.. so ans will be 4×4×3×2=96

We have four position to fill with number _ _ _ _

Now we have 5 number but we can only use any 4 number (1,2,3,4) as if we use 0 at first place then it will become a 3 digit no.

So we have (4 _ _ _)

At the second place we can use 4 number except the one which has used at first place

Similarlt at 3rd place we can use 3 number and at 4th place we can use only 2 number

So we can make (4,4,3,2) 13 such numbers(in case of without repeatation) and 19(4,5,5,5) in case of with repeatation

@himanshu its not 19

Please explain ur ans..

No repetition allowed