Inf 發問於 科學及數學數學 · 8 年前

life science

1) Blind persons read braille. A braille "cell" consists of 3 locations in each of 2 columns, or 6 locations in all. Each location may or may not have a raised dot, which can be read by touch. If at least one dot is necessary for a symbol, how many different letters, numberals, and so on can be represented by the 6 dots of a cell?

Ans: 63

2) Five visting scientists are each attached to one of two research laboratories. THe two laboratories are concerned with different areas of research and each contains 4 different departments. In how many ways can the scientists be assigned if there is to be no more than one scientist attached to any one department? In what proportion of these arrangements will two particular scientists be attached to the same laboratory?

Ans: 6720, 3/7

更新:

numberals --> numerals

更新 2:

Can someone help?? x_x

更新 3:

For Q1, thanks!

For Q2, the two particular scientists will be attached to the same laboratory, not the same department.

my approach:

5C2 * 3C2 * 4P4 * 4P1 / 6720

Is it correct?

更新 4:

但係"In how many ways can the scientists be assigned if there is to be no more than one scientist attached to any one department? In what proportion of these arrangements will two particular scientists be attached to the same laboratory?"

呢兩題唔係連住架咩?

更新 5:

唔係有一個lab只有一個scientist? 姐係果兩個particular scientists+另外兩個scientists會o係另一個lab? 咁既話點計?

3 個解答

評分
  • 8 年前
    最愛解答

    1)For each location , either have or not have a raised dot.

    Cancel the case of no dot.2^6 - 1 = 63

    2)a)5C1 * 4P1 * 4P4 ............ { [1] [ ] [ ] [ ] } { [2] [3] [4] [5] }

    + 5C2 * 4P2 * 4P3 ......... { [1] [2] [ ] [ ] } { [ ] [3] [4] [5] }

    + 5C3 * 4P3 * 4P2

    + 5C4 * 4P4 * 4P1

    = 480 + 2880 + 2880 + 480 = 6720b)假設你要和某位同房, 其餘 7 個部門只有 3 個選擇 , 機會是 3/7 。另法 :

    (同房A + 同房B) / 全部分佈

    = (4C2 + 4C2) / 8C2 = 3/7。

    2013-02-15 18:32:52 補充:

    5C1 * 4P1 * 4P4 意思是

    5個中抽一個去A房(部門分佈有4P1種) , 其餘 4 個去B房(部門分佈有4P4種)

    2013-02-15 21:04:58 補充:

    不太明你條式 , 感覺上你條式好像只有 4 , 1 分佈而沒有考慮3 , 2 分佈~

    我會咁計 :

    (同Lab4部門P2人 * 另6部門P3人) * 2Lab / 6720

    = 4P2 * 6P3 * 2 / 6720

    = 3/7

    2013-02-15 21:09:26 補充:

    我不能確定 5C2 * 3C2 * 4P4 * 4P1 / 6720 是否正確, 可否聽聽你的解析?

    2013-02-15 21:30:13 補充:

    其實真係唔駛揀...因為前提已知是那2人。

    照你的思路 :

    (3C2 * 4P4 * 4P1 + 3C1 * 4P3 * 4P2 + 3C0 * 4P2 * 4P3) * 2 / 6720

    = (288 + 864 + 288) * 2 / 6720

    = 3 / 7

    2013-02-15 21:33:13 補充:

    兩題係連住,所以應該用我們意見的2個方法較好。

    2013-02-15 21:36:54 補充:

    所有分佈 :

    Lab A : particular scientists + 0 , 1 or 2 scientists

    Lab B : ...................................2 , 1 or 0 scientists

    or

    Lab B : particular scientists + 0 , 1 or 2 scientists

    Lab A : ...................................2 , 1 or 0 scientists

    2013-02-15 21:47:04 補充:

    咪考慮晒囉 :

    (3C2 * 4P4 * 4P1 .......... 4 , 1 分佈

    + 3C1 * 4P3 * 4P2 .......... 3 , 2 分佈

    + 3C0 * 4P2 * 4P3) * 2 / 6720 ........... 2 , 3 分佈

    而 particular scientists 的 Lab 至少 2人 , 所以無 1 , 4 分佈。

    particular scientists 在另一個 Lab 時分析一樣 , 乘 2 就OK~

    2013-02-15 21:57:49 補充:

    嘿, 如果無呢句,就不能這樣算了~

    我們的計算已默認了一部門一人~

    2013-02-15 22:00:58 補充:

    我們的計算已默認了一部門至多一人~

    2013-02-15 22:05:55 補充:

    任何一個 Lab 都可有4人,每部門一人。

    因為5個人,唔會有 Lab 無人。

    部門可以無人。

    2013-02-15 22:06:54 補充:

    不用客氣~~~~~!

  • 8 年前

    Amazing!嘆為觀止,一個字【勁】^^

  • 8 年前

    我諗應該唔岩......

    5C2...係choose two particular scientists (其實係咪唔駛揀?)

    3C2...係o係淨返果三個scientists再搵兩個同果two particular scientists去同一個lab

    4P4...部門分佈

    4P1...淨低果個scientist部門分佈

    2013-02-15 21:38:40 補充:

    但兩個方法都考慮3 , 2 分佈,唔係應該只有 4 , 1 分佈?

    2013-02-15 21:53:48 補充:

    我明你既方法,但"no more than one scientist attached to any one department"呢句唔駛理?

    2013-02-15 22:04:08 補充:

    sorry!! 原來我一直睇錯題目= =

    我以為有一個lab最多一人...

    我明喇,超感謝你既回答!!!!!!!!!!

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