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# Counting challenge question #3 (hardest)

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Counting challenge question #3 (hardest) [#permalink]

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06 Dec 2008, 19:32
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This question is well beyond the GMAT (I'd say it's a 900-level question

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
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Re: Counting challenge question #3 (hardest) [#permalink]

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06 Dec 2008, 20:14
my answer is 56, although i am not happy about how i came to it..

#of 1 digit numbers: 1
For 2 digits we have (2,3),(4,1)
#of 2 digit numbers: 4 + 1 (this one is derived from adding a 0 to the 1 digit number) = 5
For 3 digits we have (1,1,3)(1,2,2)
#of 3 digit numbers: 3!/2!+3!/2!+ 1 (500) + 4*2 (adding one zero to the 2 digit numbers - 2 ways, in between or at the end) = 15
For 4 digits we have (1,1,1,2)
#of 4 digit numbers: 4!/3! + 1 (5000) + 4*3 (adding 2 zeroes to the 2 digit numbers - 3 ways, in between, at the end, or one in between and one at the end) + 6*3 (adding one 0 to the 3 digit numbers - 3 ways, after 1st digit, after 2nd digit, after 3rd digit) = 35

total = 56

took me 4+ mins - i do hope its not a GMAT Q!!
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Re: Counting challenge question #3 (hardest) [#permalink]

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06 Dec 2008, 20:50
I choose D,it also took me 8 mins to solve

positive integers less than 10,000 => the number has 4 digits

made of (5,0,0,0), (1,1,1,2), (4,1,0,0), (3,2,0,0),(3,1,1,0),(2,2,1,0)

(5,0,0,0) => 5000, 0500, 0050, 0005 => 4
Similar to (1,1,1,2) =>4

(4,1,0,0): as 1,4 can be interchangable
4100
4010
4001
0410
0401
0041
=> 6 * 2 = 12
Similar to (3,2,0,0) => 12
Similar to (3,1,1,0) => 12
Similar to (2,2,1,0) => 12

(As they have the same pattern (x,y,z,z))

Total = 4 + 4 + 12 + 12 + 12 + 12 = 56
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Re: Counting challenge question #3 (hardest) [#permalink]

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06 Dec 2008, 21:06
same here~ 56
1 digital with 0s: 5 : P(1,4)=4
2 digitals with 0s: 4+1， 2+3 : 2P(2,4)=24
3 digitals with 0s: 3+1+1， 2+2+1 : 2P(3,4)/P(2,2)=24
4 digitals: 2+1+1+1: P(4,4)/P(3,3)=4
so tatal : 4+24+24+4=56
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Re: Counting challenge question #3 (hardest) [#permalink]

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07 Dec 2008, 12:07
Nice work!
Here's my solution (copied from my workbook)
Attachments

hardest counting question.JPG [ 67.42 KiB | Viewed 2242 times ]

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Re: Counting challenge question #3 (hardest) [#permalink]

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07 Dec 2008, 22:57
brenthanneson wrote:
This question is well beyond the GMAT (I'd say it's a 900-level question

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

the possible combinations of 5's:
1: 0, 0, 0 and 5 = 4!/3! = 4
2: 0, 1, 0 and 4 = 4!/2! = 12
3: 0,0,2, and 3 = 4!/2! = 12
4: 0,1,2,and 2 = 4!/2! = 12
5: 1,1,1,and 2 = 4!/3! = 4
5: 0,1,1,and 3 = 4!/2! = 12

sum = 56
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Re: Counting challenge question #3 (hardest) [#permalink]

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24 May 2011, 09:53
I got this one also after like 6 minutes. My response would simply be this... if you were to see something this difficult on the test... just take a wild guess since you already are in the 99% rofl
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Re: Counting challenge question #3 (hardest) [#permalink]

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25 May 2011, 23:56
cool method explained by GMATTIGER.
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Re: Counting challenge question #3 (hardest)   [#permalink] 25 May 2011, 23:56
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