Quod Erat Demonstrandum



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}.


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.


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.


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


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’.


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

問該 6 題包含了老師選定的 4 題之概率。

把 3 點放入 7 個空格,每格最多放一點

7 項比賽,各項設有冠亞季 3 獎,共 21 個不同的得獎者。

為玩快活角遊戲,先隨意抽出 5 人。




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