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Mini Concept: Multinomial Theorem

with 4 comments

Some Trickery of Multinomial Theorem

x+y+z=n the number of non negative solutions of this equation will be

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

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!)= 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?

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
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

Written by Implex

October 16, 2008 at 11:45 am

Posted in Algebra, concepts

Tagged with , ,

4 Responses

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  1. Thank you …It cleared my doubt!


    October 21, 2008 at 8:12 pm

  2. yaar..phir doubt!


    then no of non negative solutions is 51c1

    here…r = 2 so, how 50c2 ?


    October 26, 2008 at 5:29 pm

    another beautiful way to solve this would be..

    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

    out of the 50 positive numbers that we have (D,2D,…) we can chose two positive numbers in 50C2 ways. And for any two positive numbers there would be only one positive number which would satisfy the condition that the sum of these three must be 0. but this includes the case in which we have selected +D and +50D and we dont have any -51D so we must reduce one case from 50C2..similiarly for case when we choose two -ves and 1 +ve..
    So final solution would be..
    50 + 50C2 – 1 + 50C2 – 1 = 2498!!
    This could have been achieved by multinomial theorem if we had used the value of 0C1 as -1 and any way 0C1 is not defined!!
    Hope I was able to put across my point


    November 11, 2008 at 3:09 pm

  4. Hey Mudit,

    shouldn’t you be subtracting 49 instead of 1 from 50C2 since any case that contains +50D and another positive number will not have a negative equivalent?


    January 18, 2009 at 12:50 pm

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