51.1999971, 41.1999989, 4.6000039, In this 3 numbers there are total 9+9+8=26 digits, if these 3 numbers are chosen randomly, what's the probability that there are 10 9's with 2 block of 4 consecutive 9's, 5 1's, 4 0's?
Let's approach this step-by-step using combinatorics:
1) First, let's consider the total number of possible arrangements:
We have 26 digits in total, and the order matters. So, the total number of arrangements is:
10^26 (as each position can be filled by any digit from 0 to 9)
2) Now, let's count the favorable arrangements:
a) We need:
- 10 nines (including two blocks of 4 consecutive nines)
- 5 ones
- 4 zeros
- 7 other digits (26 - 10 - 5 - 4 = 7)
b) Let's start by placing the two blocks of 4 consecutive nines:
We have 19 positions to place the first block (26 - 4 - 3 = 19, as we need to leave room for the second block)
Then we have 15 positions for the second block
So, there are 19 * 15 = 285 ways to place these blocks
c) We need to place 2 more nines:
We have 18 positions left, so there are C(18,2) = 153 ways to do this
d) Now, we need to place 5 ones in the remaining 16 positions:
This can be done in C(16,5) = 4368 ways
e) Next, place 4 zeros in the remaining 11 positions:
This can be done in C(11,4) = 330 ways
f) Finally, we need to fill the remaining 7 positions with the other digits:
There are 7! = 5040 ways to arrange these
g) For these last 7 digits, we can choose any digit except 0, 1, and 9:
So we have 7^7 = 823543 possibilities for what these digits could be
3) Putting it all together:
The number of favorable outcomes is:
285 * 153 * 4368 * 330 * 5040 * 823543 = 2.51654 × 10^17
4) Therefore, the probability is:
(2.51654 × 10^17) / (10^26) = 2.51654 × 10^-9
So, the probability is approximately 0.00000000251654 or about 1 in 397,371,070,190.