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This is a math problem in grade three.
1) permutation and combination? First of all, because there are only four numbers at the beginning, 1, 2, 3, 4 ..., then there is * *, and 4*4*3*2=96. 2) is a multiple of three, and the sum of digits in each position is only a multiple of three. The combination of all three multiples is: (4,3,2). 3*2+2*2+3*2= 16.3) The odd number is: 1, 3 ... so 2*2*3*2=24 4) 1, and the * at the beginning of 2,3 has 3 * (4

Beginning of 4: 40(***) has 3*2=6. Similarly, there are six 4 1(***). 420 13, 4203 1, 42 103, 42 130, so 42 130 is the 88th one. According to the above, 6 1=48+ 13, so 3203.

How many five digits can 0 1 2 3 4 make up? 4X4X3X2X 1 = 96