# 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’. 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/

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/

https://johnmayhk.wordpress.com/2013/02/07/to5e/

https://johnmayhk.wordpress.com/2009/11/26/beware-of-double-counting/