1. Let S(n,k) be the Stirling number of the second kind. There are 3! ways to permute the job positions. From defintion, the answer is 3!*S(6,3)=6*90=540.
2. To choose 3 numbers from 1,2,3,4,5,6,7 there are 10 ways:
135, 136, 137, 146, 147, 157, 246, 247, 257, 357.
Permuting the colors we get the answer 10*3!=60.
3. The answer is 84. Suppose 1，2，3，4，5，6 are the 6 positions. It is easy to list all disadjacent pairs of positions as follows:
13, 14, 15, 16, 24, 25, 26, 35, 36, 46. (1)
Again, it is easy to select a pair from the sequence (1) containing four distinct positions:
13 with 24, 25, 26, 46;
14 with 25, 26, 35, 36;
15 with 24, 26, 36, 46;
16 with 24, 25, 35;
24 with 35, 36;
25 with 36, 46;
26 with 35;
35 with 46. (2)
Counting them in (2), we get 4+4+4+3+2+2+1+1=21 pairs. For each pair, there are 2 ways to distribute the letters 1 and 2, and two ways to distribute 3 and 4. Hence the total number is
21*2*2=84.
4. I assume a mistake in the problem: the condition "any two digits have different parities" should be "any adjacent digits have different parities". Under this condition, the answer is 264.
First, by the above revised condition, either (put the letters 1,3,5,7 into the positions 1,3,5,7), or (put the letters 1,3,5,7 into the positions 2,4,6,8). It is clear that these two cases correspond to the same number of legal permutations. So we can focus on one of them, say, the first case.
Second, for any 1<=i<=8, let x_i be position of the letter i. Then the combination of (x_1,x_2) has the following 7 possibilities:
12, 32, 34, 54, 56, 76, 78. (1)
For each case, there are 3!^2=36 ways to put the letters 3,4,5,6,7,8 arbitrarily. So there are
7*36=252 (2)
ways in all.
Third, we need to find the ways among such that the letters 3 and 4 are adjacent. Suppose x_3 and x_4 are adjacent. We deal with each case in (1):
if (x_1,x_2)=(1,2), then (x_3,x_4) can be
34, 54, 56, 76, 78;
if (x_1,x_2)=(3,2), then (x_3,x_4) can be
54, 56, 76, 78;
...
if (x_1,x_2)=(7,8), then (x_3,x_4) can be
12, 32, 34, 54, 56.
In total, there are 5+4+4+4+4+4+5=30 ways of putting the letters 3 and 4 adjacently. For each of them, there are 2!^2=4 ways to put the letters 5,6,7,8. So there are
30*4=120 (3)
illegal ways in all.
Finally, by (2) and (3), we have 252-120=132 ways under the assumption that the letters 1,3,5,7 are put into the positions 1,3,5,7. Hence, the total number of legal arrangements is
132*2=264.
5. The answer is 1212. Suppose the positions to plant are respectively A,B,C,D,E,F in the clock-wise order. We write A=i to denote the plant i (1<=i<=4) is planted in the position A. Without loss of generality, we can suppose that A=1 and B=2.
If only 2 plants are used, then we have
N1=binomial(4,2)*2=12 (1)
legal ways in this case.
Suppose there are exactly 3 plants are used. Then there are 10 legal arrangements (C,D,E,F):
(1,2,1,3), (1,2,3,2), (1,3,1,2), (1,3,1,3), (1,3,2,3), (3,1,2,3), (3,1,3,2), (3,2,1,2), (3,2,1,3), (3,2,3,2).
So we have
N2:=binomial(4,3)*3!*10=240
legal ways in this case.
Suppose there are exactly 4 plants are used. It is clear that C=1 or C=3 or C=4.
If C=1 and D=2, then (E,F) is either 34 or 43. We have N3=2 ways.
If C=1 and D=3, then (E,F) is either 14, or 24 or 42 or 43. We have N4=4 ways.
If C=1 and D=4, by the symmetry of 3 and 4, we have N5=N4=4 ways.
If C=3 and D=1, then (E,F) is either 24 or 34 or 42 or 43. We have N6=4 ways.
If C=3 and D=2, then (E,F) is either 14 or 34 or 42 or 43. We have N7=4 ways.
If C=3 and D=4, then (E,F) is either 12 or 13 or 14 or 23 or 24 or 32 or 34. We have N8=7 ways.
If C=4, by symmetry of 3 and 4, we have N9=N6+N7+N8=4+4+7=15 ways.
To sum up, we have N1+N2+4!*(N3+...+N9)=12+240+4!*(2+4+4+15*2)=1212 ways in total.
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