Quod Erat Demonstrandum

2015/01/23

某數算題

Filed under: mathematics,NSS — johnmayhk @ 5:34 下午
Tags: , ,

Just reply to a F.5C student on a basic core mathematics question (on P.5.38):

There are 8 outstanding students from junior forms and 9 outstanding students from senior forms in a school this year. 5 out of these 17 students are now selected for an overseas exchange programme. Find the number of combinations of selecting at least 1 student from junior forms and 1 from senior forms.

Here is the ‘so-called’ solution from a student:

_8C_1\times _9C_1\times _{15}C_3

as the student claimed, select 1 from junior, _8C_1 ways; select 1 from senior, _9C_1 ways; then select the remaining 3 students from the remaining 15 students, _{15}C_3 ways, hence, the total number of combination should be _8C_1\times _9C_1\times _{15}C_3, right?

Sorry, it is incorrect.

The mistake is about ‘multiple counting’.

Full Spectrum Classroom

Let me represent

the 8 junior form students by {a_1,a_2,\dots ,a_8},

the 9 senior form students by {b_1,b_2,\dots ,b_9}.

CASE 1

select 1 from junior form, e.g. a_1;
select 1 from senior form, e.g. b_1;
select 3 from the remaining, e.g. a_2,b_2,b_3

This is one of the (so-called) _8C_1\times _9C_1\times _{15}C_3 cases.

CASE 2

select 1 from junior form, e.g. a_2;
select 1 from senior form, e.g. b_3;
select 3 from the remaining, e.g. a_1,b_1,b_2

This is one of the (so-called) _8C_1\times _9C_1\times _{15}C_3 cases.

CASE 3

select 1 from junior form, e.g. a_1;
select 1 from senior form, e.g. b_2;
select 3 from the remaining, e.g. a_2,b_1,b_3

This is one of the (so-called) _8C_1\times _9C_1\times _{15}C_3 cases.

However, read the three CASES above, in fact, they are the SAME! They are representing ONLY ONE SINGLE case of the selection of

{a_1,a_2,b_1,b_2,b_3}

Yes, they are NOT 3 different CASES!

That is, some cases in so-called _8C_1\times _9C_1\times _{15}C_3 cases are repeated.

To solve the question, especially the key word ‘at least’, we may think about the ‘opposite’ case (precisely, complementary event).

The ‘opposite’ of ‘At least 1 junior form and at least 1 senior form’ is ‘no junior form OR no senior form’.

Now,

number of combination of ‘no junior form’ is _9C_5 (i.e. select all 5 from 9 senior forms),

number of combination of ‘no senior form’ is _8C_5 (i.e. select all 5 from 8 junior forms),

hence the required number of combination is _{17}C_5-_9C_5-_8C_5.

~ END ~

For those who are studying simple counting in secondary school level, here are some writing in this blog, have a look if you want to

Exercise in counting balls and boxes
https://johnmayhk.wordpress.com/2011/01/07/exercise-in-counting-balls-boxes/

問該 6 題包含了老師選定的 4 題之概率。
https://johnmayhk.wordpress.com/2014/02/21/a-question-about-probability/

把 3 點放入 7 個空格,每格最多放一點
https://johnmayhk.wordpress.com/2013/07/04/boring-counting-2/

7 項比賽,各項設有冠亞季 3 獎,共 21 個不同的得獎者。
https://johnmayhk.wordpress.com/2013/03/09/a-counting-question/

為玩快活角遊戲,先隨意抽出 5 人。
https://johnmayhk.wordpress.com/2013/02/07/to5e/

(有關快樂角,小心,剛看到一文,見:https://thestandnews.com/society/%E5%B9%B3%E6%A9%9F%E6%9C%83-happy-corner-%E7%8C%B4%E5%AD%90%E5%81%B7%E6%A1%83-%E6%88%96%E6%A7%8B%E6%88%90%E6%80%A7%E9%A8%B7%E6%93%BE

小談:互斥事件@濟濟一堂
https://johnmayhk.wordpress.com/2008/04/10/applied-math-%E5%B0%8F%E8%AB%87%EF%BC%9A%E4%BA%92%E6%96%A5%E4%BA%8B%E4%BB%B6%E6%BF%9F%E6%BF%9F%E4%B8%80%E5%A0%82/

小心重覆數算
https://johnmayhk.wordpress.com/2009/11/26/beware-of-double-counting/

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