Mini Concept: Multinomial Theorem

Some Trickery of Multinomial Theorem

x+y+z=n the number of non negative solutions of this equation will be
(n+3-1)C(3-1)=(n+2)C2

if we extend this to r variables then the formula becomes (n+r-1)C(r-1)

if we remove 0, means we need only positive integral solutions to the equation then we get formula as (n-1)C(r-1)

Lets take up one example.
On the occasion of Diwali, PAPA CHIPS is offering one of five prizes with every packet( the prize is inside the packet). the prizes include a pen, pencil, a CD, a movie ticket and a small game. Banta Singh is a fan of PAPA chips and he keeps buying the chips, what is the probability that Banta Singh gets all the five prizes by buying 12 packets of PAPA chips

Solution:
Chuck the story, the question is there are 5 variables and we need the solutions to the equation
a+b+c+d+e=12 ( non negative)
and a+b+c+d+e=12 ( positive integral)
The first case comes as every packet has a prize, and those 5 are the only kinds of prizes.
Second comes from that we need each kind of prize.

so the answer is 11C4/16C4=11!12!/(7!16!)=11.10.9.8/13.14.15.16= 33/182


Lets take another example
If the sum of 101 distinct terms in arithmetic progression is zero , in how many ways can three of these terms be selected such that their sum is zero?

Solution
it is obvious that the middle term is zero
so a(51)=0
so the terms are
-50D, -49D,....,-D, 0, D, ....49D, 50D

now the sum of 3 numbers to be zero
Case 1) if we pick 0, then we have to pick one positive and one negative, which must be equal except for the sign . so 50 ways

case 2) we leave 0 and pick two positive and one negative

then xD+yD-zD=0
x+y=z
z can vary from 1 to 50
we need positive solutions to the equation
which comes 0C1+1C1+2C1...+49C1
add this it will come to 50C2

case 3 it will be same as case 2
we get 50C2

hence total is 2.50C2+50=2500