Permutation and combination

[parveen110]

Find the number of non-negative integral solutions of 2x+2y+z=10.

a. 12C2
b. 14C4
c. 11C2
d. None of these

[Brent@GMATPrepNow]

Find the number of non-negative integral solutions of 2x+2y+z=10.

a. 12C2
b. 14C4
c. 11C2
d. None of these

Hi Praveen,

This question is beyond the scope of the GMAT.
What's the source?
By the way, I noticed that you have posted only counting questions so far. Are you, by any chance, taking a Combinatorics class?

Cheers,
Brent

[parveen110]

Hi Brent,

Yes, this question is out-of-scope of GMAT. But i solve these questions only for the clarity of concepts on the topic. And of course, through this forum i may get to know the neat and purely concept-based approach to these problems because of expertise of instructors like you. I have read your posts on combinatorics. There are a lot many takeaways. But a lot of confusion remains.

And no i am not taking any class on combinatorics:) Just trying to solve on my own. This forum is really helpful.

The source of this problem is some book on aptitude that i found in a library:)

[Brent@GMATPrepNow]

Okay, fair enough.
I'm just concerned that students will see questions like these and begin worrying that they are not adequately prepared for the GMAT when, in fact, they will never see a question like this on the GMAT.

Cheers,
Brent

[GMATGuruNY]

Find the number of non-negative integral solutions of 2x+2y+z=10.

a. 12C2
b. 14C4
c. 11C2
d. None of these

This problem can be solved without high-level combinatorics.

z = 10 - 2x - 2y = even - even - even = even.
Thus, z is even.

Plug in even values for z such that 0≤z≤10 and count the number of options for x and y.
Note the following:
2x + 2y = 10-z
x+y = (10-z)/2.

Case 1: z=10
x+y = (10-10)/2 = 0.
Options for (x,y):
x=0, y=0.
Total options = 1.

Case 2: z=8
x+y = (10-8)/2 = 1.
Options for (x,y):
x=0, y=1
x=1, y=0.
Total options = 2.

Case 3: z=6
x+y = (10-6)/2 = 2.
Options for (x,y):
x=0, y=2
x=1, y=1
x=2, y=0.
Total options = 3.

Notice the PATTERN:
With every case, the number of options for (x.y) increases by 1.
Thus:
z=4 --> 4 options
z=2 --> 5 options
z=0 --> 6 options.

Adding the values in red, we get:
Total options = 1+2+3+4+5+6 = 21.

The correct answer is D.

[parveen110]

Find the number of non-negative integral solutions of 2x+2y+z=10.

a. 12C2
b. 14C4
c. 11C2
d. None of these

This problem can be solved without high-level combinatorics.

z = 10 - 2x - 2y = even - even - even = even.
Thus, z is even.

Plug in even values for z such that 0≤z≤10 and count the number of options for x and y.
Note the following:
2x + 2y = 10-z
x+y = (10-z)/2.

Case 1: z=10
x+y = (10-10)/2 = 0.
Options for (x,y):
x=0, y=0.
Total options = 1.

Case 2: z=8
x+y = (10-8)/2 = 1.
Options for (x,y):
x=0, y=1
x=1, y=0.
Total options = 2.

Case 3: z=6
x+y = (10-6)/2 = 2.
Options for (x,y):
x=0, y=2
x=1, y=1
x=2, y=0.
Total options = 3.

Notice the PATTERN:
With every case, the number of options for (x.y) increases by 1.
Thus:
z=4 --> 4 options
z=2 --> 5 options
z=0 --> 6 options.

Adding the values in red, we get:
Total options = 1+2+3+4+5+6 = 21.

The correct answer is D.

w0w!! That's really cool Mitch, thanks:)