How many ordered pairs of non-negative integers (x, y) are there which

Question

How many ordered pairs of non-negative integers (x, y) are there which satisfy the inequality  x+y≤10 ?

A. 78
B. 66
C. 55
D. 45
E. 36

ArunSharma12 · Answer

x: number of y's -> pairs
0:11
1:10
2:9 -> (2,0),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(2,7),(2,8)
..
10:1 -> (10,0)

Total number of ordered pairs = 11+10+9+8+7+6+5+4+3+2+1 = 11*12/2= 66
Ans: B

nick1816 · Answer

x+y≤10
we can write the above inequality as

x+y+k=10, where k is non-negative integer.

Total number of ordered solutions = 12C2 = 66

sayan640 · Answer

How come it's 12 C 2  ?   MartyMurray  can you please help on this question ?

Kinshook · Answer

How many ordered pairs of non-negative integers (x, y) are there which satisfy the inequality  x+y≤10 ?

x = 0; y = {0,1,2,....10}
x = 1; y = {0,1,2,....9}
x = 2; y = {0,1,2,.....8}

x=10; y = {0}

Total ordered pairs = 11 + 10 + .... + 1 = 11*12/2 = 66

IMO B

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