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
Thursday, October 16, 2008
Mini Concept: Multinomial Theorem
Wednesday, October 15, 2008
Problem Of The Week 53
How many ways can a size k + 1 subset with maximum element m + 1 can be created from the given set S={1,2,3,.....,n+1} ?
Friday, October 10, 2008
Problem Of The Week 52
Suppose that
and
are positive numbers for which
,
what is the value of
?
what is the value of
Problem Of The Week 51
Let P(x)=ax^4 +bx^3+cx^2+dx+e be a polynomial with all integer coefficients and a=1. If√2+√5 is one of the roots of P(x)=0 , which of the following can be the value of |(b+c+d+e)|?
1) 103
2) 89
3) 63
4) 23
5) 5
Thursday, October 9, 2008
Problem Of The Week 50
Find the number of unordered triplets (x,y,z) of positive integers such that x³+y³+z³=2008
1) 0
2) 1
3) 2
4) 3
5) None of these
1) 0
2) 1
3) 2
4) 3
5) None of these
Labels:
Algebra,
Number Theory,
Number Thoery,
Problem of the week
Problem Of The Week 49
Find the maximum area that can be bound by four line segments of length 1, 2, 3 and 4. (you are not allowed to break a segment, you may join two :) )
1) 6√2
2) 2√6
3) 4√6
4) 3√2
5) None of these
1) 6√2
2) 2√6
3) 4√6
4) 3√2
5) None of these
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