IIT JEE , AIEEE Forum - Mathematics , Algebra

lokeshsardana

Find the number of integral solutions of the equatiion

2x + y + z = 20

where x,y,z >= 0

actually I have solved this question but with a lengthy method.....

but I want to know other methods too to solve this question......

so, plz tell all the methods you know to solve this problem.

note: pls xplain clearly each and every step.......

all good efforts will be rated...

akhil_o

now x,y,z>=0, we have to find solns of
2x+y+z=20
2x term is even...et us take it to be a
now a varies from 0 to 20...y and z vary accordingly
for a=0, solns of x+y=20, a=2, x+y=18 and so on till a=20...
so required solution is sumation of all these cases

for a=0
y+Z=20, it can be done by the ball-wall method...it can be done such that 20 balls are placed...we place a wall in one of the 21 gas so that they automatically get divided into 2 groups

a=0...21C1=21
a=2 19C1=19
and so on

so required soln
= 21+19+17+....3+1
=11+10(11)
=121

sory evry one....extremly stupid typing error...
thanks Greatdreams and sniper

the.sniper

ext{We have the equation} 2x+y+z=20 ext{consider the following cases:} ext{Case1: }x=0 ext{we have }y+z=20

now we have to divide 20 in two parts in such a way that each part gets 0 or more

this can be done in²¹C₁ ways

similarly on considering cases when x=1 ,x=2 ...x=10we get ¹⁹C1 ,¹⁷C₁......1C₁

adding number of solutions of all cases we get 21+19+17+.......1 = sum of first 11 odd numbers= 11²=121

Greatdreams

2x + y + z = 20

So let x = r

So you have y + z = 20 - 2r

0

r

10 ( as it is given that x,y,z

0)

So we have Number of integral solutions

= No of ways to distribute (20 - 2r) objects among 2 persons

=^{20 - 2r + 2 - 1} C_{2 - 1}

[ No of ways to distribute n objects among r persons = ^n+r-1 C _r-1 ]

=^{21 - 2r}C ₁ = 21 - 2r

So no of reqd solutions
= _{[r =0]}

¹⁰ (21 - 2r) =21 + 19 + 17 + 15 + 13 +...+ 3+ 1

= 121 ways

Akhil please check your answer......

sboosy

ax₁+bx₂+cx₃= n

where x₁ x₂ x₃ are integers

a_i <= x_i <=b_i (i= 1 , 2 , 3)

coeff of xⁿ in [ (x^a)^{a_{1 +}} (x^a )^a₁+1.....upto x power a power b₁ )] [ .....] [ ....]

in the second bracket instead of a put b ...that is coeff of x₂ and instead of a1 ...put a₂ to b₂

similarly third bracket u put c _instead of a and change substcripts to 3 from 2

according to given sum

a=2 0<=x₁<= 10

b=1 0<=x₂<=20

c=1 0<=x₃<=20

so second and third bracket become [ 1+ x + x²+ ....x²⁰] each

first bracket

[1+x²+x⁴+ ...x²⁰]

so question nothing but finding coeff of x power 20

in [1+x²+x⁴+ ...x²⁰] [ 1+ x + x²+ ....x²⁰][ 1+ x + x²+ ....x²⁰]

this is the method

lokeshsardana

good work guys........

@ akhil ........sorry dude ....ur answer is not correct...

@ sniper and greatdreams............I appreciate ur work ...but both of u provided same method.......which I used initially to solve this problem.........

@ sboosy ...... r u sure u r getting same answer( i.e.121) from this method........well this method seems to be more lengthy........so, I didn't tried to find out the answer.................

anyways...........once again thanx to all of u....4 ur solutions....

rates to all of u.........

sboosy

i think the method gives the same answer ..121

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