linear inequalities

(i) x < 2k -1 及 x > 3k + 2
無解 當 x < x
即 x < 2k -1 <= 3k + 2 < x
k >= -3
等於時也不行，因為左右都是 strictly 不等。你對。
(ii) non-negative (非負) 即可零或任何正數
先考慮簡單情況
n = 1 (x,y) = (1,0), (0,1), (0,0) 三個choice
n = 2 (x,y) = (2,0), (1,1), (1,0), (0,2), (0,1), (0,0) 六個choice
n = 3 (x,y) = (3,0), (2,1),(2,0),(1,2),(1,1),(1,0),(0,3),(0,2),(0,1),(0,0) 十個 choice
n = 4 (x,y) = (4,0),(3,1),(3,0),(2,2),(2,1),(2,0),(1,3),(1,2),(1,1),(1,0),(0,4),(0,3),(0,2),(0,1),(0,0) 十五個 choice

其實上面有條 nCr 係用得arm的，但你未必學左nCr, 同埋 nCr 意思其實即係咁樣，係list out 簡單情況之後，再分析就可以睇得出的

你留唔留意到我排列D choice ge 特定次序? 咁樣咪數得晒lor,
首先, fix n. x 由大數到細, 有 n+1 個 選擇, n,n-1,n-2,....2,1,0
x = n 時, y 只可以是0,
x = n-1 時, y 可以是 1 或 0,
...
x = 0 時, y 可以有 n+1 個選項
所以答案係 1+2+3+...+(n+1)
這個在中四會學公式,公式係咁樣黎的
Let Sum = 1+2+3+..+(n+1)
Sum + Sum(掉轉次序) = 1+(n+1) + 2+n + 3+(n-1) +.... + (n+1)+1
= n+1 個 n+2
therefore, 1 個 Sum = (n+1)*(n+2) / 2

1

2x - 1 < 4k - 3

2x < 4k -2

x < 2k -1.

2x - 10(k 1) > 5k -3x

5x > 5k 10k 10

x > 3k 2.

As the inequalities to have no solution,

Then
3k 2 > 2k -1
Therefore,

k > -3.
2

if x and y are non-negtive integers,
the number of solutions that satisfy ' x y <= n '
= (n 2)C2.......(the combination of solutions)
i.e.

the number of solutions that satisfy ' x y <= n '
is (n 1)(n 2) / 2 .

Q.1
2x - 1 < 4k - 3
2x < 4k -2
x < 2k -1.
2x - 10(k+1) > 5k -3x
5x > 5k + 10k + 10
x > 3k + 2.
For the inequalities to have no solution, the 2 possible regions must not overlap.
That is 3k+ 2 > 2k -1 .Therefore,
k > -3.
Q.2
Seems that x and y can be any positive integer since n is also any positive integer.