parveen110 wrote: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.
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