I assume the first one, that is, at least two numbers have the same tail, and vice versa.
The possible types of 49 people on duty are A (7 7,49) = 49 * 48 * 47 * ... * 43.
Divide these 49 numbers into 5 rows 10 columns, as follows (0 is added neatly).
0 - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9
10 - 1 1 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19
20 - 2 1 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29
30 - 3 1 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39
40 - 4 1 - 42 - 43 - 44 - 45 - 46 - 47 - 48 - 49
Simply calculate, seven numbers have different endings and are divided into two categories.
One: When the first column is not included, just take seven columns out of the other nine columns and arrange them.
Is A (7, 9), each column has five possibilities, so multiply it by 5 7.
There are (7,9) * 5 7 possibilities for this situation.
Two: When the first column exists, only six columns need to be taken out from the other nine columns for arrangement, A(6.9)
Multiplied by 4 * 5 6, a * * has a (6 6,9) * 4 * 5 6 possibilities.
So the possible types of different tails are A (6,9) * 4 * 56+A (7,9) * 57.
Probability of the same tail
p= 1-[a(6,9)*4*5^6+a(7,9)*5^7/a(7,49)
(As a result, you should figure it out yourself. MS is older, I will be lazy when I steal. )
How many tail numbers do you draw a week?
This is the biggest problem. What does this mean?
If the four mantissas of 10, 20, 30, 40 are the same, how many mantissas are 40, 20,1,2 1, 3 1, 22, 23? Conversely, 40, 20, 1 1, 2 1, 3 1, 22, 23 count as three.
10,20,30,40 is one?
I won't forget it here. The above is just a personal opinion. LZ should think about it for reference only.