some permutation & combination qs help pls!!!

[JS]

am 數學白痴 need some detail explainations for the following question. many thx!!!

1, n個中國人，n個美國人，n個日本人排成一列，同國籍不相鄰，排法有幾種？

2, AB兩隊比賽足球最先贏3局者為勝沒有和局試求比賽所有可能發生的情形有幾種

3, 投兩個骰子 求點數和出現3的倍數的機率是多少


感激不盡

[hsiaoj]

am 數學白痴 need some detail explainations for the following question. many thx!!!

1, n個中國人，n個美國人，n個日本人排成一列，同國籍不相鄰，排法有幾種？

2, AB兩隊比賽足球最先贏3局者為勝沒有和局試求比賽所有可能發生的情形有幾種

3, 投兩個骰子 求點數和出現3的倍數的機率是多少


感激不盡

用我非常久遠的赫哲記憶試試看 錯了大家幫忙糾正
1. 2* C 2n+1取n
想像先排中國人和美國人 有"中美中美..."或是 "美中美中..."2種方法
排完後插入日本人 (總共會有2n+1個位置可以排日本人)
從這些位置選n個放進那些人 即是答案

2.20 畫樹枝圖就OK

3. 1/3
硬要列式子 就一個一個數吧
(1,2) (2,1)
(1,5) (2,4) (3,3) (4,2) (5,1)
(3,6) (4,5) (5,4) (6,3)
(6,6)
機率= 12/36=1/3


不過 這真是gmat嗎???
gmat有這麼麻煩的題目嗎

[JS]

thank u hsiaoj for your kindness.
but what is "樹枝圖"?
also can anyone help me with another question-

五封寫好的信，任意放入五個寫好收信人及地址的信封內，每個信封僅裝一封信，求下列各種狀況的機率？
(1)
至少有一封信放對了信封。
(2)
沒有一封信放對了信封。

someone said
恰]放對一封信的情況-->5x9=45
[恰]放對兩封信的情況-->10x2=20
[恰]放對三封信的情況-->10x1=10
[恰]放對四封信的情況(放對4封,第五封一定也對)
-->1

is it correct and why?

[hsiaoj]

thank u hsiaoj for your kindness.
but what is "樹枝圖"?
also can anyone help me with another question-

恕刪!

"樹枝圖" 好難解釋

你就把 A 獲勝寫A, B獲勝寫B
然後去延伸
這個版面沒半法畫圖 我剛才嘗試失敗 ><
自己在紙上試試看吧 加油!

[micht]

Tree diagram長這樣

[hsiaoj]

Tree diagram長這樣

micht 真是太棒了

[JS]

thx a lot u super girls.
How about the other question?
五封寫好的信，任意放入五個寫好收信人及地址的信封內，每個信封僅裝一封信，求下列各種狀況的機率？
(1)
至少有一封信放對了信封。
(2)
沒有一封信放對了信封。

I know that 沒有一封信放對的事件有
5！- 5˙4！+ 10˙3！- 10˙2！+ 5˙1！- 1˙0！ = 44
But can anyone explain this to me in detail??
I have looked everywhere for the answer but still no luck.........
help pls!!

[liwuu]

Tree diagram長這樣

呵呵...這個圖看起來好像是外國教科書上面的圖!!
micht真利害...什麼東西都有!!

[liwuu]

someone said
恰]放對一封信的情況-->5x9=45
[恰]放對兩封信的情況-->10x2=20
[恰]放對三封信的情況-->10x1=10
[恰]放對四封信的情況(放對4封,第五封一定也對)
-->1

所以全部的情形-上面情形加總=沒有一封放對的情形
而至少有一封放對的情形為上面情形的加總

[JS]

someone said
恰]放對一封信的情況-->5x9=45
[恰]放對兩封信的情況-->10x2=20
[恰]放對三封信的情況-->10x1=10
[恰]放對四封信的情況(放對4封,第五封一定也對)
-->1

所以全部的情形-上面情形加總=沒有一封放對的情形
而至少有一封放對的情形為上面情形的加總

hi liwuu, ya, I know that part, what I dont understand is why
恰]放對一封信的情況-->5x9=45 and why
恰]放對兩封信的情況-->10x2=20 and so on..

I thought the total possible number of getting one letter right is
C(5,1) . 4! and getting 2 letters right is
C(5,2). 3!

[liwuu]

嗯嗯...這個問題我想不出可以直接用什麼樣的算式來計算表達(請高手指教)
我也是只能用樹枝圖的方式來計算
以放對一封信的情況來看
若假設A放對,其他皆放錯的情況下
則可能為ACBED,ADBEC,AEBDC,ACEBD,AEDBC,ADEBC,ACDEB,ADECB,AEDCB等九組
所以只有一封信放對的總情形為C(5,1)*9

[JS]

Finally, after days of searching, I find the answer that makes sense to me. Here it comes...(just in case that you are wondering as well)

If we let A be the event that letter A is in ist correct envelope and
similarly B is the event that letter B is in its correct envelope,
then

P(A) = 1/5 and

P(A and B) = 1/5 * 1/4

Now use the inclusion-exclusion principle to get probability that A or
B or C .... or E are correctly placed.

P(A or B or C .... or E) = P(A) + P(B) + P(C) + P(D) + P(E)
- P(A and B) - P(B and C) -....
+ P(A and B and C) + P(B and C and D) + ....
- P(A and B and C and D) - P(...) -......
+ P(A and B and C and D and E)

= 5*(1/5)
- 5C2*(1/5)(1/4)
+ 5C3*(1/5)(1/4)(1/3)
- 5C4*(1/5)(1/4)(1/3)(1/2)
+ (1/5)(1/4)(1/3)(1/2)(1/1)

5*4 1 5*4*3 1 5*4*3*2 1 1
= 1 - --- * --- + ----- * ----- - --------- * ------- + ---------
1*2 5*4 1*2*3 5*4*3 1*2*3*4 5*4*3*2 5*4*3*2*1

= 1 - 1/2! + 1/3! - 1/4! + 1/5!

This is the probability that at least one letter is correctly placed.

The chance that none is correctly placed is 1 - (above result), or

= 1 - 1 + 1/2! - 1/3! + 1/4! - 1/5!

= 1/2! - 1/3! + 1/4! - 1/5!

With n letters and envelopes, the probability that none is correctly
placed is:

= 1/2! - 1/3! + 1/4! - 1/5! + ....... + (-1)^n 1/n!

or use this method (for getting all the letters in the wrong envelops)
5! - 5C1 x 4! + 5C2 x 3! - 5C3 x 2! + 5C4 x 1! - 5C5 x 0! = 44