How many positive integers less than 10,000 are there in which sum of digits equals 5? (a) 31 (b) 51 (c) 56 (d) 62 (e) 93

Hi, Let abcd be the number. We need to find the number of non-negative integral solutions of a+b+c+d = 5 It is (5+4-1)C4-1 = 8C3 = 56 Hence, C If x1+x2+..+xk = n, then the number of non-negative solutions is (n+k-1)C(k-1).

How did you get that it is a combination problem? I could not! Does this kind of problems need "combination" approach?

Ozlemg wrote:How did you get that it is a combination problem? I could not! Does this kind of problems need "combination" approach? This is not a combination problem directly. But, by solving we end up with this formula. This will save time, else you need to consider cases: 1)count 1 digit numbers - only one(5) 2)count 2 digit numbers - five(14,23,32,41,50) 3)count 3 digit numbers ... 4)count 3 digit numbers ... 5)count 3 digit numbers ...

Ozlemg wrote:How many positive integers less than 10,000 are there in which sum of digits equals 5? (a) 31 (b) 51 (c) 56 (d) 62 (e) 93 Another approach is to use the separator method. Check here: https://www.beatthegmat.com/experts-any- ... 82307.html

GMATGuruNY wrote: Another approach is to use the separator method. Check here: https://www.beatthegmat.com/experts-any- ... 82307.html Exactly. This separator method gives the formula (n+k-1)C(k-1).

the sum should be 5.maximum it can be a 4 digit number. In the following cases the sum is 5. 1004 --> In 12 ways it can be represented ( 4!/2!) 1103---> 12 ways 2003---> 12 ways 2102---> 12 ways 1112----> 4 ways (4!/3!) 5000---> 4 ways In total 56 ways