How well can you add?

Simple solution using stars and bars: We can image any combination of 8 digits that adds up to 4 as 8 boxes (represented by 7 "bars" as separators between boxes) into which 4 stars are distributed. For example 30001000 -> xxx||||x|||

In order for such combination to form an 8 digit number, it must not start with 0. Therefore we put the first star in the first box. Now we just need to distribute remaining 3 stars into 8 boxes. There are ${{7 + 3} \choose 3} = 120$ ways of doing that.

Interpret the 8 digit number as a string of 8 numbers $\in \{0,1,2,3,4\}$ . Note that we cannot have a digit strictly larger than 4, or else the sum of the digits will evidently be at least 5.

Denote this string of numbers as : $\{a_1,a_2, \cdots , a_8 \}$ , where $a_1$ denotes the units digit of our number.

Now, for our number to the made up of 8 digits, we require : $a_8 \neq 0$ . Hence, we proceed by a case-to-case analysis :

• $Case \ 1 : a_8 = 4$

Here, we need to set $a_i \equiv 0, \quad \forall i=1, \cdots,7$

Number of possibilities : 1

• $Case \ 2 : a_8 = 3$

For the sum to be 4, we need one of the $a_i$ 's to be 1 and the others 0.

Number of possibilities : 7 (7 choices to choose which $a_i =1, \ \ i \in \{1, \cdots,7\}$ )

• $Case \ 3 : a_8 = 2$

Here we have 2 sub cases : either we set only 1 other digit to be 2 and the others 0, or we set 2 digits to be 1.

Number of possibilities : $7+C^2_7 = 7+21 = 28$

• $Case \ 4 : a_8 = 1$

3 sub cases here : exactly digit is 3, one digit is 1 and another 2 or three digits are 1.

Number of possibilities : $7 + 2*C^2_7+C^3_7 = 7+42+35 = 84$

Hence, total number is : $1+7+28+84 = \boxed{120}$

Number of combinations whose sum is 4 are: 1111,22,31,40,112

1111: we have to make 8 digit number by 1111.this can be done in 7C3 ways =35ways

22: eight digit number can be made in 7C1 ways=7 .

31: first fix 3 to the largest place value and arrange '1' at seven places by 7C1 ways=7ways. Next fix 1 to the largest place value and arrange'3' at remaining seven places by 7C1 ways=7 ways

40: eight digit number can be made in only one way=1

112: fix'1'to the largest place value then the remaining seven places can be filled in 7P2 ways=42 Fix'2'to the largest place value then the remaining seven places can be filled in 7C2 ways=21ways

TOTAL NUMBER OF COMBINATIONS TO FORM EIGHT DIGIT NUMBER WHOSE SUM IS 4 IS=35+7+7+7+1+42+21=120

Suppose the number is a 1 a 2 a 3 ... a 8 .The possible values of a_1 are 1, 2, 3, 4. We consider these four cases.

If a_1 = 4 then all other digits are 0 (since sum of digits is 4). Hence there is only 1 such number.

If a_1 = 3 then exactly one of the other 7 digits is 1. Hence there are 7 such numbers (depending on where the digit ‘1’ is).

If $latex a_1 = 2 $ then sum of the other seven digits is 2.

Hence we compute the number of non negative integer solutions of $latex a 2 + … + a 8 = 2 $ .

This equals \binom {6+2}{2} = 28

If a_1 = 1 then sum of the other seven digits is 3.

Hence we compute the number of non negative integer solutions of a 2 + ... + a 8 = 3

This equals \binom {6+3}{3} = 84

Hence the answer is 120.