Question

高二数学排列组合问题，求详解

1）6名大学生去3个单位应聘，若每个单位至少录用其中一人，则不同的录用情况个数为？
2）有红、黄、蓝三种颜色的球各7个，每种颜色的7个球分别标号1、2、3、4、5、6、7，从中任取3个标号不同的球，这3个颜色互不相同且所标数字互不相邻的取法个数为？
3）由两个1、两个2、一个3、一个4六个数字组成6位数，要求相同数字不能相邻，则这样的6位数个数为？
4）用1、2、3、4、5、6、7、8组成8位数（无重复数字），要求任意两个数字的奇偶性不同，且1、2相邻，3、4不相邻，这样的8位数个数为？
5）将一个正六边形按对角线平分成6个部分，每一部分栽种一种植物，要求相邻两部分的植物不同，现有4种不同植物可供选择，则共有多少种不同的栽种方案？

Answer 1

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.

Answer 2

1（1）6个人都被录取，C(6,2)×A(3,3)+C(6,3)×C(3,2)×C(1,1)×A(3,3)+C(6,2)×C(4,2)×C(2,2)×A(3,3)/A(3,3) =90+360+90=540(种）
（2）有5个人被录取，则有：[C(5,3)×A(3,3)+C(5,2)×C(3,2)×C(1,1)×A(3,3)/A(2,2) ]×C(6,5)=(60+90)×6=900（种）
（3）有4个人被录取，则有：C(4,2)×A(3,3)×A(6,4)=36×15=540（种）
（4）3个人被录取，则有： A(3,3)×C(6,3)=6×20=120（种）
∴共有 540+900+540+120=2100（种）
2 如果最小数为1，只可能为135 136 137 146 147 157 六种，同理若为最小为2，只能是246 247 257 为3则是 357 4或以上不可能，共10种全排乘A33 =60个

5
以A、C、E（相间）栽种植物情况作为分类标准：
①A、C、E栽种同一种植物，有4种栽法；B、D、F各有3种栽法，
∴ 共有 4×3×3×3＝108 种栽法。
②A、C、E栽种两种植物,有 种栽法（ 是4种植物中选出2
种， 是A、C、E3个区域中选出2个区域栽种同一种植物， 是
选出的2种植物排列），B、D、F共有3×2×2 种栽法（注：若A、C栽种同一种植物，则B有
3 种栽法，D、F各有2种栽法），

③A、C、E栽种3种植物，有 种栽法；B、D、F各有2种栽法，
∴ 共有 ×2×2×2＝192 种栽法。
综合①、②、③，共有 108+432+192=732种栽法。

三四题楼主自己努力……鄙人不会……

Answer 3

D3
84种，用部分合条件问题排除法。
D4,576种