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Fantezi Cebir => Fantezi Cebir => Konuyu başlatan: stuart clark - Ekim 15, 2012, 04:21:53 ös

Başlık: no. of integer solution
Gönderen: stuart clark - Ekim 15, 2012, 04:21:53 ös

Number of Integer solution of the equation |x|+|y|+|z| = 10

Başlık: Ynt: no. of integer solution
Gönderen: Lokman Gökçe - Ekim 16, 2012, 08:39:27 ös

we see that for n > 0 integers,

number of solutions |x| + |y| = n is 4n.

If |z| = 0 then z = 0 and |x| + |y| = 10. So, number of solution = 1.4.10=40

If |z| = 1 then z = ±1 and |x| + |y| = 9. So, number of solution = 2.4.9

If |z| = 2 then z = ±2 and |x| + |y| = 8. So, number of solution = 2.4.8

... etc

If |z| = 9 then z = ±9 and |x| + |y| = 1. So, number of solution = 2.4.1
If |z| = 10 then z = ±10 and |x| + |y| = 0. So, number of solution = 2.1=2

total = 40 + 2.4(9+8+...+1) + 2 = 402

Başlık: Ynt: no. of integer solution
Gönderen: stuart clark - Ekim 21, 2012, 05:29:47 öö

Thanks Admin Got it

Başlık: Ynt: no. of integer solution
Gönderen: geo - Ekim 21, 2012, 12:10:16 ös

Alıntı yapılan: stuart clark - Ekim 15, 2012, 04:21:53 ös

Number of Integer solution of the equation |x|+|y|+|z| = 10

We will use combination with repetition (http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition).
Suppose |x|,|y|,|z| are all positive. In that case, if we rewrite |x|=a+1, |y|=b+1, |z|=c+1, we will get a+b+c=7.
There are C(7+3-1, 7)= 36 ways to distribute 7 objects into 3 boxes.
But x=-a-1, y=-b-1, z=-c-1 are also solutions. So there are 2³x36=288 ways.

Suppose two of |x|, |y|, |z| are 0. These two can be selected in C(3, 2) ways.
So there are 2xC(3, 2) = 6 ways.

Suppose only one of |x|, |y|, |z| is 0. This one can be selected in C(3, 1) ways.
Let |x|=0 and |y|=b+1, |z|=c+1.
We will get b+c = 8.
There are C(8+2-1,8) = 9 ways to distribute 8 objects into 2 boxes.
Since y, z can be negative, there are 2²xC(3, 1)xC(8+2-1,8) =108 ways.

In total, there are 288+108+6 = 402 ways.


More general case:
|x₁|+|x₂|+...+|x_r| = n

Suppose k of them are non-zero.
Let |x_i| = a_i + 1 where i is a positive integer less than k+1.
Σa_i+1 = n
Σa_i = n-k

n-k objects can be distributed k boxes in C((n-k)+k -1, n-k) = C(n-1, n-k)  ways.
The non-zero ones can be selected in C(r, k) ways.
Each none-zero can be negative.

So if k of r numbers are non-zero,
z_k =  2^k.C(r, k).C(n-1, n-k)

Total number of ways is
Σ^r_k=1 2^k.C(r, k).C(n-1, n-k)

For our case:
r=3, n=10
Σ³_k=1 2^k.C(3, k).C(10-1, 10-k)
= 2¹.C(3, 1).C(10-1, 10-1) + 2².C(3, 2).C(10-1, 10-2)[/b] + 2³.C(3, 3).C(10-1, 10-3)
= 2.C(3,1).C(9,9) + 4.C(3,2).C(9,8) + 8.C(3,3).C(9,7)
= 6 + 108 + 288
= 402

Başlık: Ynt: no. of integer solution
Gönderen: stuart clark - Aralık 11, 2012, 05:48:27 öö

Thanks Bosbeles