AC and BD are the same, and because there are only three issues, ABCD must be available, so it can only be 1)AC, B_, D _；; 2) DB，A_，C _； Two ways. But AC, BD, _ _ will be repeated; According to the repeatable course arrangement order.
1. 1. 1) ? The first AC/CA:? Two kinds? , b (in the second paragraph): (16-3abc) 2+(16-3) x 3 (bbx, BXB, XBB)? +1(BBB)-13x13x3+1= 209, d (three segments): (16-3acd) 3+6 (.
1. 1.2)? The first AC/CA:? Two kinds? , d (in the second paragraph): (16-3adc) 2+(16-3) x3 (ddx, DXD, XDD)? +1(DDD) ditto = 209, b (three segments): (16-3abc) 3+6 (bbxx, bxbxbx, bxbx, xbbx, xxbb, xbxbxb) x132 (xx).
BDX ( 13) x 3(BDX，BXD，xbd)+6 (BBD，DDB...) = 45; BD is calculated in three times: BDXX (13 2) x 6 (BDXX, BXDX, BXXD, DBXX, DXBX)+6 (bbdx, ddbx ...) x13+6 (bbdd ...)+4 (bbbd). *** 45 x 3264，209 x 1 108
In other words, the ways of AC in the first paragraph are:
2 x 209 x 3264+2 x 209 x 3264-(45 x 3264+209 x 1 108)
1.2. 1) Phase II AC:( 16-4 ABCD)x 6(ACX, CAX...)+6 (AAC, ACA, CAA, ACC, CAC, CCA)? =78, b comes first: (16-3abc) x2 (bx, XB)+ 1 (bb) = 27, d 3:? 3264
1.2.2)? The second AC: ditto =78, the first D:27, B: 3? 3264
1.2.3) BD:2(BD, DB) in the first repetition; BD is repeated in three sections: same as1.1.3)1108.
In other words, the ways of AC in the second paragraph are:
78 x 27 x 3264+78 x 27 x 3264-(78 x？ 2 x 3264 + 78 x 27 x 1 108)
1.3. 1)? The third AC: (16-4 ABCD) 2x6 (acxx ...)+24 (aacx, ACCX…… ...)? = 1038, b comes first: same as 1.2. 1) 27, and d comes last: same as1.1) 209.
1.3.2)? The third AC: 1038, the first D:27 and the second B:? 209
1.3.3) BD:2(BD, DB) in the first repetition; The second paragraph repeats BD: same as 1. 1.3) 45.
In other words, the ways of AC in the third paragraph are:
1038 x 27 x 209+ 1038 x 27 x 209-( 1038 x？ 2 x 209+ 1038 x 27 x 45)
So is BD. I don't think the answer is reasonable. There should be conditions that cannot be repeated.