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positive integers less than 10,000

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by krishnasty » Fri Jul 22, 2011 10:38 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) 934

OA : C
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by cans » Sat Jul 23, 2011 12:34 am
# of one digit numbers = 1 (only 5 satisfies)
# of 2 digit numbers = 5 (14,41,23,32,50)
# of 3 digit numbers = 15(122 (*3) ,131 (*3),410 (*4),230 (*4),500 (*1))
# of 4 digit numbers =35 (5000 (*1), 4100 (*6), 3110 (*9), 3200 (*6), 2210(*9), 2111(*4))
OA C
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by GMATGuruNY » Sat Jul 23, 2011 1:36 am
I posted a solution here:

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by Ozlemg » Sat Jul 23, 2011 3:51 am
cans wrote:# of one digit numbers = 1 (only 5 satisfies)
# of 2 digit numbers = 5 (14,41,23,32,50)
# of 3 digit numbers = 15(122 (*3) ,131 (*3),410 (*4),230 (*4),500 (*1))
# of 4 digit numbers =35 (5000 (*1), 4100 (*6), 3110 (*9), 3200 (*6), 2210(*9), 2111(*4))
OA C
Hi Cans

Thank you for your help!
I did not understand the bold numbers?
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by Ozlemg » Sat Jul 23, 2011 4:01 am
krishnasty 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) 934

OA : C
This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is-->n+r+1Cr-1

8C3 is 56.

Hence, C
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by winniethepooh » Sat Jul 23, 2011 4:12 am
That is the number of ways those digits can be written to give a sum of five:
122 can be arranged in 3!/2! ways. = 3 ways

410 can be arrangen in 6(3!/0!) ways in all = 6 ways - 2 ways in which 0 appears as the first digit as already considered earlier in two digits(14). Hence, 4 ways.

Similarly others.
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by Brent@GMATPrepNow » Sat Jul 23, 2011 8:19 am
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Aside: We can extend this solution to conclude that the number of integers less than 1,000,000 in which the sum of the digits equals 8 will be 13C5

Cheers,
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