哪位高手帮小弟我写个函数方法

谁帮我写个函数方法
最近在研究彩票的问题，有一个关于复式数的分解问题始终在困扰我。

例子

PHP code

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/**
比如4个号码分解成3个号码和2个号码
4码：[1,2,3,4] 一组号码
分解成3码后:[1,2,3] [1,2,4] [1,3,4] [2,3,4] 四组三位数的号码
分解成2码后:[1,2] [1,3] [1,4] [2,3] [2,4] [3,4] 六组二位数的号码
不考虑位置

又如5个号码分解成4个号码
4码：[1,2,3,4,5] 一组号码
分解成4码后:[1,2,3,4] [1,2,3,5] [1,2,4,5] [1,3,4,5] [2,3,4,5]四组四位数的号码
分解成3码后:[1,2,3] [1,2,4] [1,2,5] [1,3,4] [1,3,5] [1,4,5] [2,3,4] [2,3,5] [2,4,5] [3,4,5] 十组三位数的号码
分解成2码后:[1,2] [1,3] [1,4] [1,5] [2,3] [2,4] [2,5] [3,4] [3,5] [4,5] 十组二位数的
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大家应该看出规律了吧
谁能帮我写个方法
array lottery(array $num, int $i){
}
@param $num 是指一组数字
@param $i 是指分解的码数
返回结果数组

如
$arr1 = lottery(array(1,2,3,4,5), 4);
$arr2 = lottery(array(1,2,3,4,5), 3);
此时
$arr1 的值应该是一个数组
[1,2,3,4] [1,2,3,5] [1,2,4,5] [1,3,4,5] [2,3,4,5]
而$arr2 的值也是一个数组
分解成3码后:[1,2,3] [1,2,4] [1,2,5] [1,3,4] [1,3,5] [1,4,5] [2,3,4] [2,3,5] [2,4,5] [3,4,5]




这个问题我反复研究过，貌似挺难的。。过年了，给66分，祝答题者六六大顺
*/

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PHP code

function Combination($arr, $size = 1) { 
    $len = count($arr); 
    $max = pow(2,$len) - pow(2,$len-$size);
    $min = pow(2,$size)-1; 
 
    $r_arr = array(); 
    for ($i=$min; $i';
print_r($r); 
echo '

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'; /* Array ( [0] => Array ( [0] => 1 [1] => 2 [2] => 3 ) [1] => Array ( [0] => 1 [1] => 2 [2] => 4 ) [2] => Array ( [0] => 1 [1] => 3 [2] => 4 ) [3] => Array ( [0] => 2 [1] => 3 [2] => 4 ) [4] => Array ( [0] => 1 [1] => 2 [2] => 5 ) [5] => Array ( [0] => 1 [1] => 3 [2] => 5 ) [6] => Array ( [0] => 2 [1] => 3 [2] => 5 ) [7] => Array ( [0] => 1 [1] => 4 [2] => 5 ) [8] => Array ( [0] => 2 [1] => 4 [2] => 5 ) [9] => Array ( [0] => 3 [1] => 4 [2] => 5 ) ) */