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Re: If x, y and z are non-negative integers such that x < y < z, [#permalink]
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GMATPrepNow wrote:
If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78


0≤x<y<z
x+y+z=11
x=0, y=1, z=11-1=10: average (10+1)=11/2=5.5 so y={1 to 5}={1,2,3,4,5}=5
x=1, y=2, z=11-3=8: average (8+2)=10/2=5 so y={2 to 4}={2,3,4}=3
x=2, y=3, z=11-5=6: average (6+3)=9/2=4.5 so y={3 to 4}={3,4}=2
x=3, y=4, z=11-7=4: invalid

total solutions: 5+3+2=10

Answer (B)
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If x, y and z are non-negative integers such that x < y < z, [#permalink]
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BrentGMATPrepNow wrote:
If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78



The number of ways 11 can be divided into 3 buckets is:

13!/11!2! = 78

Eliminate those in which all 3 are not distinct.

Since 11 isn't divisible by 3, this means those where two are the same and one distinct:

0 0 11
1 1 9
2 2 7
3 3 5
4 4 3
5 5 1

Each of these can be arranged 3 ways for a total of:

18

So a preliminary count is:

78-18 = 60

Since the remaining 60 comprise sets of distinct numbers arranged 3! = 6 ways and the question is identifying one arrangement of x<y<z, the number of solutions is:

60/6 = 10

Posted from my mobile device
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Re: If x, y and z are non-negative integers such that x < y < z, [#permalink]
Regor60 wrote:
BrentGMATPrepNow wrote:
If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78


The number of ways 11 can be divided into 3 buckets is:

13!/11!2! = 78

Eliminate those in which all 3 are not distinct.

Since 11 isn't divisible by 3, this means those where two are the same and one distinct:

0 0 11
1 1 9
2 2 7
3 3 5
4 4 3
5 5 1

Each of these can be arranged 3 ways for a total of:

18

So a preliminary count is:

78-18 = 60

Since the remaining 60 comprise sets of distinct numbers arranged 3! = 6 ways and the question is identifying one arrangement of x<y<z, the number of solutions is:

60/6 = 10

Posted from my mobile device

­This method is extremely time-saving compared to the listing of numbers. :thumbsup:  :thumbsup:
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Re: If x, y and z are non-negative integers such that x < y < z, [#permalink]
@BrentGMATPrepNowAsked: If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?
Number of distinct solutions = {(0,1,10),(0,2,9),(0,3,8),(0,4,7),(0,5,6),(1,2,8),(1,3,7),(1,4,6),(2,3,6),(2,4,5)}

IMO B
Quote:
 

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Re: If x, y and z are non-negative integers such that x < y < z, [#permalink]
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