How many of the five numbers from 1 to 33 add up to 106? Five numbers can't be repeated. The order is indefinite

There are many kinds. If five numbers can't have one repetition at all, there are the following four kinds: (in the case of all natural numbers)
1:6+16+26+28+30=106
2:1+9+31+32+33=106
3:7+23+24+25+27=106
4:14+20+21+22+29=106

How many groups of numbers are there?
If you can, teach me the fastest way to combine it

I don't know where you come from. I calculated it with computer programming sequence. If six numbers are different, there can be 19760 groups. I didn't find a simple algorithm

How to add any seven numbers from 1 to 33 when the sum equals 107
If you have any information, please help me

If all the numbers are not reused and have nothing to do with the position of the numbers, there are 64526 kinds of numbers. The minimum starting point is 12353233 and the maximum ending point is 1231415161819. If it is related to the position of the numbers, there are 64526 × 7! = 64526 × 5040 = 325211040 kinds of numbers

Seven rows and seven columns from 1 to 49 are equal to one number, how to arrange them

The simplest is this method:
Row 1 in the middle of the first row (column 4), and all the numbers in the back row are arranged to the upper right. If you reach the boundary, turn from the boundary to the other side (for example, 2 turns to the seventh row and fifth column). If you encounter the number that has been filled, move down one space (for example, 7 is in the second row and third column, and 8 will touch 7, so move down to the third row and third column)
Just fill it in

Math 1-9 add and subtract equals 10

1+9

How does it hold that if you add 1 to 9 and subtract 100

FOUND:123-45-67+89=100
FOUND:12-3-4+5-6+7+89=100
FOUND:12+3+4+5-6-7+89=100
FOUND:123+4-5+67-89=100
FOUND:1+2+3-4+5+6+78+9=100
FOUND:12+3-4+5+67+8+9=100
FOUND:1+23-4+56+7+8+9=100
FOUND:1+2+34-5+67-8+9=100
FOUND:1+23-4+5+6+78-9=100
FOUND:123+45-67+8-9=100
FOUND:123-4-5-6-7+8-9=100

4.8.2.4.6.9.5, adding or subtracting is equal to ten,

4-8-2-4+6+9+5=10

Why is it that any number from 1 to 9 multiplied by 3 plus 3 and then multiplied by 3 equals 9 when the number of digits and tens add up

Let this number be a
Then (3a + 3) × 3 = 9A + 9 = 10A + (9-A)
Ten is above and the number is a
The number is (9-A)
So a + 10-A = 9

Why is it that if any number from 1 to 9 is multiplied by three plus three and then multiplied by three, the sum of the single digit and the ten digit of the result is equal to 9?

It's all nine times

From 1 to 9, if you add 5 digits, you will get 24? (you can have repeated digits in an equation)
There can be repeated numbers in a formula, such as: 4 + 2 + 2 + 1 + 1

If so, that's a lot, such as: 1 + 1 + 1 + 1 + 20, 1 + 1 + 1 + 2 + 19

How many of the five numbers from 1 to 33 add up to 106? Five numbers can't be repeated. The order is indefinite