Let me briefly talk about a concept that i remember studying long back.Most of the logic would be excerpts from Higher Algebra by Bernard Shaw

We know that most of probability questions can be easily cracked if we have a strong grip on Permutation and combination.Let me present you all with something interesting. I'll be dealing with "Distribution" of similar and dissimilar things

Concept 1: Distributing Similar things among people where each recipient receives at least 1 thing

----------------------------------------------------------------------------------------------

Imagine you have 10 similar tennis balls which you need distribute it to 3 friends and each of the fella receives at least 1 ball.

how would you approach it???

Let us do it figuratively.Imagine the 10 balls

O O O O O O O O O O ( remember there are 9 spaces between 10 balls)

if you consider 2 sticks ! !, how many ways can u place them in the 9 spaces between the 10 balls. yeah right 9C2 ways.. would u believe , thats the answer. for example consider O O ! O O O ! O O O O O . this is one of the 9C2 ways where the first guy gets 2 balls, 2 nd guy gets 3 and the last gets 5. you just need to place (r-1) sticks in the spaces between the n balls and that's like selecting r-1 spaces between the n balls.

so the answer is n-1Cr-1 n= number of balls r= number of recepients

I can rephrase the same question

Find the number of solution to the equation

X + Y + Z = 10 where each X, Y and Z integers and are >0

consider X =Friend 1 Y= Friend 2 and Z= Friend 3 and the number 10 on the other side of the eqn as the 10 similar balls

hence u already know the solution 9C 2

Concept 2 : Distributing Similar things among people where each recipient may receive no balls at all

----------------------------------------------------------------------------------------------

Imagine the same old story where you have 10 similar tennis balls which you need distribute it to 5 friends. However now 1 or more of them may receive 0 balls

Again imagine the balls lying beautifully in the imagination of your brain

O O O O O O O O O O .

Now as you know for 5 friends you need to consider 4 sticks ! ! ! !. Now that any of the friend may receive 0 balls also, we cannot consider the space between the balls. So what do you think.. let me figuratively sample u some solution

one of the solution O ! O ! O O ! O O O ! O O O ( FRIEND 1= 1 BALL, FRIEND 2= 1 BALL FRIEND 3= 2 BALLS FRIEND 4= 3 BALLS FRIEND 5 = 3BALLS)

ANOTHER SOLUTION ! O O O ! O O O ! O O O O ! ( FRIEND 1 = 0 BALLS , FRIEND 2= 3 BALLS , FRIEND 3= 3 BALLS , FRIEND 4 = 4 BALLS FRIEND 5 = 0)

I hope by now you would have guessed the solution is nothing but arranging 14 things of which 10 are of one kind and 4 are of one kind

that is !14/(!10*!4) . this is nothing but n+r-1 C n where n is the number of balls and r is the number of recipients

I can again rephrase the same question

Find the number of solution to the equation

X + Y + Z = 10 where each X, Y and Z integers and are >=0

consider X =Friend 1 Y= Friend 2 and Z= Friend 3 and the number 10 on the other side of the eqn as the 10 similar balls

hence u already know the solution 14C4

I can present some more topics. this will be based on the condition that the readers find it interesting and helpful

Hi Thank you very much for these information. I found it very interesting. However, I need your help on your conclusion: X+Y+Z=10, should the solution be 12C2 instead of 14C4? I'm confused.