saketk wrote:

Brent@GMATPrepNow wrote:This one is well past the 800-level, but there are a lot of clever people on this site who will, I imagine, accept the challenge.

How many positive integers less than 10,000 are there in which the sum of the digits equals 5?
(A) 31
(B) 51
(C) 56
(D) 62
(E) 93
Answer: C

Try to resist examining each case and counting all of the possibilities. To discourage this strategy I could have asked "How many positive integers less than 1,000,000 are there in which the sum of the digits equals 8?"

Hi Brent -- Got this link from your latest post..

This is how I did it (though I used bit of a counting here)




------------------------------------------------
We can make numbers from ---0,1,2,3,4,5 ONLY
------------------------------------------------
1 digit number

only 1 = 5

TOTAL = 1 way
-----------------------------------------------
2 digit numbers

possible sets (2,3) = 2*1 = 2 ways
if (4,1) then 2*1 = 2 ways
if (5,0) then = 1 way

TOTAL = 5 ways
------------------------------------------
3 digit numbers

possible sets (5,0,0) then =1 way

if (4,0,1)
2*2*1 = 4 ways

if one is 3 then 2 sets possible
(3,0,2) = 2*2*1 = 4 ways

(3,1,1) = 3*2*1/2 = 3 ways

if one is 2 then set can be (2,2,1) = 3*2*1/2 = 3 ways ... NOTE- we have already used a set of (2,3,0)

TOTAL = 1+4+4+3+3 = 15 ways

---------------------------------------------------

4 digits

possible sets (1,2,2,0) (1,1,1,2) (1,3,1,0) (3,2,0,0) (4,1,0,0) (5000)

_ _ _ _

3*3*2*1 /2 = 9 ways

4*3*2*1/6 = 4 ways

3*3*2*1/2 = 9 ways

2*3*2*1/2 = 6 ways

2*3*2*!/2 = 6 ways

1 way

respectively


TOTAL = 18+16+1 = 35 ways
--------------------------------------------------------
GRAND TOTAL = 35+21 = 56

sana.noor wrote:using the inserting stick and seperator technique

(5+3)!/5!.3! = 56...C is the answer

is my approach to answer this question right?