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How many positive integers less than 10,000 are there in whi

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by rakeshd347 » Sun Oct 06, 2013 8:06 pm
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

OA is C

I have a confusion about this question. When we are solving with the separator method. We are considering 5 numbers to be distributed among 4 people. So we have 5 numbers and 3 separators. This way will give 8!/5!3!....but my question is aren't we considering 0,1,2,3,4,5 so we have 6 digits to distribute to 4 people. Can you please spot where I am wrong. I know We don't have to consider 0 and only 1,2,3,4,5....but 5000 is also the digit which have sum equal to 5.
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by vipulgoyal » Sun Oct 06, 2013 8:33 pm
I have come across with brute force method, which took me 10 minuts and impractial in exam scenarios

one digit No - 5 -- 1
two digit no - 14,41,23,32,50 -- 5
three digit - 113(3 ways) , 221(3 ways), 500(one way), 401(4 ways), 302(4 ways) -- 15
four digit - 1112(4 ways), 1220(9 ways),1310(9ways),1004(6ways),2003(6 ways), 5000 (one way) -- 35 ways

35+15+5+1= 56

I didnt get what you are trying to ask??
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by theCodeToGMAT » Sun Oct 06, 2013 8:40 pm
Digits : _ _ _ _

We need to distribute : 1 1 1 1 1 = 5

Using Seperator Method:

_ | _ | _ | _

Hence, 8!/5!3! = 8*7 = 56

Answer [spoiler]{C}[/spoiler]
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by rakeshd347 » Sun Oct 06, 2013 8:58 pm
theCodeToGMAT wrote:Digits : _ _ _ _

We need to distribute : 1 1 1 1 1 = 5

Using Seperator Method:

_ | _ | _ | _

Hence, 8!/5!3! = 8*7 = 56

Answer [spoiler]{C}[/spoiler]
Hi Rahul,

This is where I am getting confused. You have used four _ and three separators I so the total number is 7. So shouldn't it be 7!/4!3!
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by Brent@GMATPrepNow » Sun Oct 06, 2013 9:07 pm
rakeshd347 wrote: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
My solution can be found here: https://www.beatthegmat.com/very-tricky- ... 25349.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by rakeshd347 » Sun Oct 06, 2013 9:09 pm
theCodeToGMAT wrote:Digits : _ _ _ _

We need to distribute : 1 1 1 1 1 = 5

Using Seperator Method:

_ | _ | _ | _

Hence, 8!/5!3! = 8*7 = 56

Answer [spoiler]{C}[/spoiler]
Rahul I think I have got the answer.

5 Different chocolate has to distributed among 4 kids....this question is same as your question as we are considering 4 digit numbers.
so we need 3 separator.
OOOOO and III
if its 0005 then it will represent....III00000
if its 1211 then it will represent....0I00I00

So it counts to 8 so the answer is 8!/3!5!
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by theCodeToGMAT » Sun Oct 06, 2013 9:19 pm
rakeshd347 wrote:
theCodeToGMAT wrote:Digits : _ _ _ _

We need to distribute : 1 1 1 1 1 = 5

Using Seperator Method:

_ | _ | _ | _

Hence, 8!/5!3! = 8*7 = 56
Answer [spoiler]{C}[/spoiler]
Rahul I think I have got the answer.

5 Different chocolate has to distributed among 4 kids....this question is same as your question as we are considering 4 digit numbers.
so we need 3 separator.
OOOOO and III
if its 0005 then it will represent....III00000
if its 1211 then it will represent....0I00I00

So it counts to 8 so the answer is 8!/3!5!
Bingo!
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