counting problem 1

Sppose 3 people receive x , y , z coins respectively .

It is clear that both x,y and z are non-negative integers .

# of ways is equivalent to # of integral solutions of

x + y + z = 6 (*) , where x,y,z >= 0

Now , let x'=x+1 , y'=y+1 , z'=z+1 , then x',y',z' are positive integers satisfying

x' + y' + z' = 9 ( ** ) . Clearly , # of solution of (*) = # of solution of (**)

Finally , to count # of solution of (**) , we need to put two L's to separate

the 0's by 3 groups .

0_0_0_0_0_0_0_0_0

There are 8 "_" , thus 8C2 ways to do so .

Therefore , # of ways to distribute the coins

= # of positive integral solution of (**)

= 8C2

Here is a method using "combination" :

Firstly, put the six identical coins (denoted as "O") in a row as shown :
_O_O_O_O_O_O_

Then put 2 separators (denoted as "|") into the 7 spaces (denoted as"_") to divide the 6 coins into 3 groups.
The 3 persons A, B and C get the group of coins from left to right.
The number of coins that each person get is ranged from 0 to 6.

The first case is to put the 2 separators into 2 different spaces
(e.g. | O_O_O_O_O_O |)
Number of ways in this case = 7C2

The second case is to put the 2 separators into the same space.
(e.g. _O_O_O_O || O_O_)
Number of ways in this case = 7C1

Total number of ways
= 7C2 + 7C1
= 7!/5!2! + 7!/6!1!
= 21 + 7
= 28

(The given answer 8C2 = 28)

The question is similar to :
How many number of ways to distribute 9 identical coins to 3 different people, if each of them should have at least 1 coin.
So the answer is 8C2

There are 8! different ways.

Reason:

The question is equivalent to the following question:

If we try to put 2 sticks to separate 6 balls into 3 groups, how many ways can we do so?

O|OOO|OO

So,

The answer

= The no. of ways to arrange the 2 sticks and 6 balls

= The no. of ways to arrange 8 objects

= 8!



Wish it helps.