Fill in 4 spaces, leaving 1 spaces. Error category 2

Fill in 3 blanks and leave 2 blanks. All error types A (2,4)-C (1,2) × 2-C (2,2) = 7.

Type A (3 3,5)-c (1,3) × 7-c (2,3 3) × 2-C (3 3,3) = 32, in which 2 spaces are filled and 3 spaces remain.

The filling pair of 1 is empty, and the remaining four empty ones are all error types A (4 4,6)-C (1,4) × 32-C (2,4 4) × 7-C (3 3,4 )× 2-C (4,4) =/kloc.

5 Zero-total-wrong species A (5 5,7)-C (1,5) ×181-C (2,5 )× 32-C (3,5 )× 7-C (4,5 )× 2-C (.

......................., did you find the recursive call here? ......................

therefore

There are 65,438+0 kinds, and there are five vacancies.

C (4,5) × 2 =10, with four vacancies.

C (3,5) × 7 = 70 kinds, with only three vacancies.

C (2,5) × 32 = 320 species, with just two blanks.

Just fill in the empty species C (1, 5) × 18 1 = 905.

Species of 0+0.5 × 12 14 = 12 14.

Let's calculate the probability of correctly answering n space, and the denominator is a (5,7). Then, use the basic mathematical expectation algorithm and do it yourself. There was no resistance. The final answer is 1.43, which means 10/7.

This is an internal complex analysis. If it makes sense, divide the upstairs 5 by 7 and multiply it by 2 ~