Combinations

[baladon99]

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

Can any one help to solve this problem ?

Thanks in advance,
Bala

[havok]

So I'll try to type and figure out this problem at the same time. It has to be less than 10,000, so it's going to be a number with one, two, three, or four digits.

* One Digits * 5 (1 Total)
* Two Digits * 50, 41, 32, 23, 14 (6 total)
* Three Digits * 500, 410, 401, 320, 302, 311, 230, 203, 221, 212, 140, 104, 113, 131, 122 (21 Total)
* Four Digits * 5000, 4100, 4010, 4001, 3200, 3020, 3002, 3110, 3101, 3011, 2300, 2030, 2003, 2120, 2102, 2210, 2201, 2120, 2102, 2111, 1400, 1040, 1004, 1130, 1103, 1220, 1202, 1310, 1301. (51 Total)

B. 51 (wrong), there should be a short cut to a correct answer?

[vvsmart]

Only 1 non-zero digit - 4
5, 50,500,5000 - 4

two non-zero digits - 24

23 - 2 ( 23 & 32)
230 - 4 (230, 203, 320, 302)
2300 - 6 (2003, 2030, 2300, 3200, 3020, 3002)

Total = 12

or calculate overall as 4!/2! = 12 ( find combinations for 4 digits where 2 are repeating 2, 3, 0, 0)

Similarly 12 options for 1,4,0,0

3 non-zero digits - 24

1,2,2,0
Total = 4!/2! = 12

1,3,1,0
Total = 4!/2! = 12

4 non-zero digits = 4

1,1,1,2
Total = 4!/3! = 4

Overall total = 56