If R (B1, B2, B3)

If R (B1, B2, B3)

incorrect.
For example:
b1=(0,0,0)^T,b2=(0,0,0)^T,b3=(0,0,1)^T
a1=(1,0,0)^T,a2=(0,1,0)^T,a3=(0,0,0)^T
Obviously, R (B1, B2, B3) = 1

Arrange 1,2,3,4,5,6.37 into a row A1, A2, A3,. A37, where a1 = 37, A2 = 1 and make a1 + A2 + a3 + A4 + A5 +. + a (k) divisible by a (K + 1) (k = 1,2,3,4,5,6,7.36). Find 1) a37,2) A3 (the process should be detailed)

Because a1 + A2 = 38, it needs to divide A3, so A3 = 2 or 19. If A3 = 2, then A37 = 19, because a1 + A2 + a3 + +A36 = 37 * 38 / 2-19 = 36 * 19, can divide A37. If A3 = 19, then A37 = 2, because a1 + A2 + a3 + +A36 = 37 * 38 / 2-2 = 2 * 2 * 3 * 3 * 3 * 13, can divide A37. So, the above two cases are true, that is, 3

Arrange 1,2,3,4,5,6.37 into a row A1, A2, A3,. A37, where a1 = 37, A2 = 1 and make a1 + A2 + a3 + A4 + A5 +. + A37 divisible by a (K + 1) (k = 1,2,3,4,5,6,7.36). Find 1) a37,2) A3

a37=19
a3=2
The title is wrong, it should be a1 + A2 + +AK can be divided by a (K + 1), the solution of this problem mainly uses the decomposition of 38

In the complete permutation a1a2a3a4a5 of 1, 2, 3, 4, 5, the number of permutations satisfying A1 < A2, A2 > A3, A3 < A4, A4 > A5 is ()
A. 10B. 12C. 14D. 16

When the middle digit is 1, choose two of the remaining four digits and put them in the first two places. The order of the last two digits is determined. There are six results of C42 = 6. When the middle digit is 2, there are six results of the above situation. When the middle digit is 3, 4 and 5 can only be put in the second and