Permutation and Combination

a)( ) B ( ) B ( ) B ( ) B ( )
4! ways for 4 'B'.
4 'G' from 5 '( )' , 5P4 ways.
4! (5P4) = 2880 required ways.
b)3a + 2b + c = 18
{
a + b + c = 7
(0 ≤ a , b , c ≤ 7)⇒ b + 2c = 3 and hence (b = 1 , c = 1 , a = 5) or (b = 3 , c = 0 , a = 4).(7C1)(6C1)(5C5) + (7C3)(4C0) = (7)(6)(1) + (35)(1) = 77 required ways.
ci)6! ways for all cases , 5! for the number of ways of cases that '0' is the 1st digit.
6! - 5! = 600 required numbers.cii)35??? : 4P3 numbers
4???? : 5P4 numbers
5???? : 5P4 numbers4P3 + 5P4 + 5P4 = 24 + 120 + 120 = 264 required numbers.ciii)Group A : 0 , 3
Group B : 1 , 4
Group C : 2 , 5Choosing one number from each group : 2³ = 8 ways.
For each choosed case , there are 3! = 6 permutation , (6)(8) = 48 numbers ,
(or 3×2 × 2×2 × 1×2 = 48 numbers) including 0?? , (2!)2² = 8 numbers,
So there are 48 - 8 = 40 required numbers.

(a)由於女孩子不可排在一起，一定要被男隔開，所以ANS是:
4P4 x 5P4 = 2880

(b)要7條問題總分加起來剛好是18，每題得分的分佈有以下兩個可能:
3+3+3+3+2+2+2 = 18 OR 3+3+3+3+3+2+1 = 18
可能性1的排序 = 7! / (4!)(3!) =35
可能性2的排序 = 7! / (5!)(1!)(1!) = 42
總數= 35 + 42 = 77

(ci) 5 x 5 x 4 x 3 x 2 x 1 (6位數不可以是0開頭) =600

(cii) 1 x 1 x 4 x 3 x 2 (開頭是3的可能) + 2 x 5 x 4 x 3 x 2 (開頭是4 or 5的可能) = 264

(ciii) 要被3整除，3位的數字加上來一定要是3的倍數，因此可能性組合有:
1,0,2 --> 3
1,0,5 --> 6 or 2,0,4 --> 6 or 1,2,3 --> 6
4,0,5 --> 9 or 3,1,5 --> 9 or 4,2,3 --> 9
3,4,5--> 12
再把以上的可能性組合排序並相加，因此答案是:
(2x2x1) x 4 + (3!) x 4 = 40

以上乃是本人自己算出來的，可能有誤，如果有更快或正確的方法，歡迎其他數學高手作出指教。

謝謝支持～

我終於可以排第一～

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