Sheldon ross example 5b,5c

Question

Q.1) Ten children are to be divided into an A team and a B team of 5 each. The A team
will play in one league and the B team in another. How many different divisions are
possible?

Q.2) In order to play a game of basketball, 10 children at a playground divide themselves
into two teams of 5 each. How many different divisions are possible?

Answer 1

let reduce this into small size problem

let there are 4 people named them as A,B,C and D, we want to divide them as size of 2

Formula is C(4,2) = 6

what are those ?

i) ( A , B )

ii) ( A , C )

iii) ( A , D )

iv) ( B , C )

v) ( B , D )

vi) ( C , D )

 

1) As per First question, Teams are Numbered ===> the first team can be any one of the choice, after choosing that team there is no choice for choosing

 like if i choose team A = (A,B) then i should choose team B = (C,D)

∴ Total Possibilities = 6 = C(4,2)

 

2) in the second question, Teams are Not Numbered, means

if you are saying choose (A,B) this is equivalent to choose (C,D) as team

if you are saying choose (A,C) this is equivalent to choose (B,D) as team

if you are saying choose (A,D) this is equivalent to choose (B,C) as team

∴ Total Possibilities = 3 = $\frac{6}{2}$ = $\frac{ C(4,2)}{2}$

 

For your question

A) C(10,5)

B) $\frac{ C(10,5)}{2}$

Comments

@shaikh bro really good explanation

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@Shaik Masthan  what if for ques2, instead of dividing into equal number of players, we divide into  6 and 4 players. Then ans will 10!/(6!4!) rather than ( 10!/(6!4!) )/2!. Because now the two teams are distinguishable. Right?

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Yes @ayush.5