If x, y and z are non-negative integers such that x

Question

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

vishalkazone · Accepted Answer

X, Y, Z Is non negative integer, and x+y+z=11, and x<y<z
We can fix value of x and calculate distinct solutions
 Case-1 
X=0 
Y can have 1,2,3,4,5 (Total 5 values) to reach x+y+z =11.
For each value of y z will have a fix value. So total 5 distinct solutions.
 Case-2 
 x=1
Y can be 2,3,4, for each value, z will have a fix value ( 6,7, and 8)
Total 3 distinct solutions in this case.
 Case -3 
X=2
Y can be 3,4 for each value, z will have a fix value ( 5,6,)
Total 2 distinct solutions in this case.
 Case -4 
X=3
No value of y can satisfy x+y+z=11, and x<y<z.

Total
5+3+2=10
B.

BrentGMATPrepNow · Answer

I created this question to show that there can be times when the best (i.e., fastest) way to solve a counting question is by  listing and counting

How do we know when it's not a bad idea to use listing and counting? 
The answer choices will tell us (ALWAYS scan the answer choices before beginning any answer choices)
Here, the answer choices are reasonably small, so listing and counting shouldn't take long. 
ASIDE: Yes, 78 (answer choice E) is pretty big. However, if you start listing possible outcomes and you eventually list  more than  22 outcomes (answer choice D), you can stop because the answer must be E.

Okay, let's list possible outcomes in a systematic way. 
We'll list outcomes as follows:  x, y, z
 
Since x is the smallest value. 
Let's list the outcomes in which  x = 0 . We get: 
 0, 1, 10 
0, 2, 9
0, 3, 8
0, 4, 7
0, 5, 6

Now, the outcomes in which  x = 1 . We get: 
 1, 2, 8
1, 3, 7
1, 4, 6

Now, the outcomes in which  x = 2 . We get: 
 2, 3, 6
2, 4, 5
 
Now, the outcomes in which  x = 3 . We get: 
3, 4...hmmm, this won't work.

So, we're done listing!
Count the  outcomes  to get a total of 10

Answer: B

Cheers, 
Brent

exc4libur · Answer

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)

Regor60 · Answer

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

samarpan.g28 · Answer

This method is extremely time-saving compared to the listing of numbers. :thumbsup:

Kinshook · Answer

@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

If x, y and z are non-negative integers such that x < y < z,

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